25.3 - Formation of a Differential Equation When General Solution is Given

In the previous section, we completed a discussion on order and degree of a differential equation. In this section, we will see formation of a differential equation when general solution is given.

Some basic details can be understood by analyzing some examples.

Example 1
This can be written in 5 steps:
1. Consider the equation $\small{x^2~+~y^2~=~r^2}$.
• We know that, it is the equation of the circle with radius r and center at the origin.
• By giving different values for 'r', we can obtain infinite number of concentric circles. Some of those circles are shown in the fig.25.6 below:

Fig.25.6

   ♦ The green circle has a radius of 2 units.
   ♦ The red circle has a radius of 3 units.
   ♦ The yellow circle has a radius of 5 units.
so on . . .

• All circles have a common center, which is O. We can write:
The equation $\small{x^2~+~y^2~=~r^2}$ represents a family of concentric circles with center at O 

2. The above equation contains a constant 'r'. On many occasions that we encounter in science and engineering, we will want an equation which does not contain constants.
• In the present case, let us try to find such an equation. For that, we differentiate the equation as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x^2~+~y^2}    & {~=~}    &{r^2}    \\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{2x~+~2y \frac{dy}{dx}}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x~+~y \frac{dy}{dx}}    & {~=~}    &{0}    \\
\end{array}}$

3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{x~+~y \frac{dy}{dx}~=~0}$ represents the family of concentric circles with center at O.

4. Take any member of the family. Say $\small{x^2~+~y^2~=~8^2}$
• This member will be a solution of the differential equation that we wrote in step (3) above. Let us check. It can be checked in two steps:

(i) Find the derivative from the given solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x^2~+~y^2}    & {~=~}    &{8^2}    \\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{2x~+~2y \frac{dy}{dx}}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x~+~y \frac{dy}{dx}}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{(-1)x}{y}}    \\
\end{array}}$

(ii) Substitute the above derivative in the differential equation:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x~+~y \frac{dy}{dx}}    & {~=~}    &{0}    \\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{x~+~y \left[\frac{(-1)x}{y} \right]}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x~+~(-x)}    & {~=~}    &{0}    \\
\end{array}}$
Which is true.

• So $\small{x^2~+~y^2~=~8^2}$ is indeed a solution.

5. A summary of the analysis can be written in three steps:
(i) We were given an algebraic equation. It represents a family of concentric circles. It has an arbitrary constant 'r'.
(ii) Based on that algebraic equation, we obtained a differential equation. It does not have the arbitrary constant 'r'.
(iii) Any member of the family mentioned in (i), is a solution of the differential equation mentioned in (ii).

Example 2
This can be written in 5 steps:

1. Consider the equation $\small{x^2~+~y^2~+~2x~-~4y~+~4~=~0}$.
This can be rearranged as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x^2~+~y^2~+~2x~-~4y~+~4}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{x^2~+~2x~+~y^2~-~4y~+~4}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x^2~+~2x~+~1~+~y^2~-~4y~+~4~-~1}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{(x+1)^2~+~(y-2)^2~-~1}    & {~=~}    &{0}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{(x+1)^2~+~(y-2)^2}    & {~=~}    &{1}    \\
\end{array}}$

• We know that, it is the equation of the circle with radius 1 unit and center at (−1,2).
• We can write it in a general form as:
$\small{(x+1)^2~+~(y-2)^2~=~r^2}$
• We know that, it is the equation of the circle with radius 'r' units and center at (−1,2).
• By giving different values for 'r', we can obtain infinite number of concentric circles. Some of those circles are shown in the fig.25.7 below:

Fig.25.7

   ♦ The green circle has a radius of 2 units.
   ♦ The red circle has a radius of 3 units.
   ♦ The yellow circle has a radius of 5 units.
so on . . .

• All circles have a common center, which is (−1,2). We can write:
The equation $\small{(x+1)^2~+~(y-2)^2~=~r^2}$ represents a family of concentric circles with center at (−1,2)
 
2. The above equation contains an arbitrary constant 'r'. On many occasions that we encounter in science and engineering, we will want an equation which does not contain arbitrary constants.

• In the present case, let us try to find such an equation. For that, we differentiate the equation as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{(x+1)^2~+~(y-2)^2}    & {~=~}    &{r^2}    \\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{2(x+1)~+~2(y-2) \frac{dy}{dx}}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x+1~+~(y-2) \frac{dy}{dx}}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{x+1}{2-y}~(y \ne 2)}    \\
\end{array}}$

3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{\frac{dy}{dx}~=~\frac{x+1}{2-y}~(y \ne 2)}$ represents the family of concentric circles with center at (−1,2) .

4. Take any member of the family. Say $\small{(x+1)^2~+~(y-2)^2~=~5^2}$
• This member will be a solution of the differential equation that we wrote in step (3) above. Let us check. It can be checked in two steps:

(i) Find the derivative from the given solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{(x+1)^2~+~(y-2)^2}    & {~=~}    &{5^2}    \\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{2(x+1)~+~2(y-2) \frac{dy}{dx}}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{(x+1)~+~(y-2) \frac{dy}{dx}}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{x+1}{2-y}~(y \ne 2)}    \\
\end{array}}$

(ii) Substitute the above derivative in the differential equation:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{x+1}{2-y}~(y \ne 2)}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{x+1}{2-y}~(y \ne 2)}    & {~=~}    &{\frac{x+1}{2-y}~(y \ne 2)}    \\
\end{array}}$

Which is true.

• So $\small{(x+1)^2~+~(y-2)^2~=~5^2}$ is indeed a solution.

5. A summary of the analysis can be written in three steps:
(i) We were given an algebraic equation. It represents a family of concentric circles. It has an arbitrary constant 'r'.
(ii) Based on that algebraic equation, we obtained a differential equation. It does not have the arbitrary constant 'r'.
(iii) Any member of the family mentioned in (i), is a solution of the differential equation mentioned in (ii).

Example 3
This can be written in 5 steps:

1. Consider the equation $\small{y~=~3 x^2~+~\rm{C}}$.

• We know that, it is the equation of a parabola.
• By giving different values for 'C', we can obtain infinite number of parabolas which belong to a family. Some of those parabolas are shown in the fig.25.8 below:

Fig.25.8

   ♦ The green parabola has C = 0.
   ♦ The red parabola has C = 5
   ♦ The yellow parabola has C = −3
so on . . .

• All parabolas are symmetrical about the y-axis. We can write:
The equation $\small{y~=~3 x^2~+~\rm{C}}$ represents a family of parabolas having axis along positive y-axis.
 
2. The above equation contains an arbitrary constant 'C'. On many occasions that we encounter in science and engineering, we will want an equation which does not contain arbitrary constants.

• In the present case, let us try to find such an equation. For that, we differentiate the equation as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{3 x^2~+~\rm{C}}    \\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{dy}{dx}}    & {~=~}    &{6x}    \\
\end{array}}$

3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{\frac{dy}{dx}~=~6x}$ represents a family of parabolas having axis along +ve y-axis.

4. Take any member of the family. Say $\small{y~=~3 x^2~+~4}$
• This member will be a solution of the differential equation that we wrote in step (3) above. Let us check. It can be checked in two steps:

(i) Find the derivative from the given solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{3 x^2~+~4}    \\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{dy}{dx}}    & {~=~}    &{6x~+~0}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{dy}{dx}}    & {~=~}    &{6x}    \\
\end{array}}$

(ii) Substitute the above derivative in the differential equation:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{6x}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{6x}    & {~=~}    &{6x}    \\
\end{array}}$

Which is true.

• So $\small{y~=~3 x^2~+~4}$ is indeed a solution.

5. A summary of the analysis can be written in three steps:
(i) We were given an algebraic equation. It represents a family of parabolas. It has an arbitrary constant 'C'.
(ii) Based on that algebraic equation, we obtained a differential equation. It does not have the arbitrary constant 'C'.
(iii) Any member of the family mentioned in (i), is a solution of the differential equation mentioned in (ii).

Example 4
This can be written in 5 steps:

1. Consider the equation $\small{y~=~a x^2~+~\rm{C}}$.

• We know that, it is the equation of a parabola.
• By giving different values for 'a' and 'C', we can obtain infinite number of parabolas which belong to a family. Some of those parabolas are shown in the fig.25.9 below:

Fig.25.9

   ♦ The green parabola has a = 3 and C = 0.
   ♦ The red parabola has a = 0.5 and C = 4.
   ♦ The yellow parabola has a = 0.2 and C = −2.
so on . . .

• All parabolas are symmetrical about the y-axis. We can write:
The equation $\small{y~=~a x^2~+~\rm{C}}$ represents a family of parabolas having axis along +ve y-axis.
 
2. The above equation contains two arbitrary constants 'a' and 'C'. On many occasions that we encounter in science and engineering, we will want an equation which does not contain arbitrary constants.

• In the present case, let us try to find such an equation. For that, we differentiate the equation as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{a x^2~+~\rm{C}}    \\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{dy}{dx}}    & {~=~}    &{2ax}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{d^2y}{dx^2}}    & {~=~}    &{2a}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\frac{d^2y}{dx^2}}    & {~=~}    &{\frac{dy/dx}{x}}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{x\frac{d^2y}{dx^2}}    & {~=~}    &{\frac{dy}{dx}}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{x\frac{d^2y}{dx^2}~-~\frac{dy}{dx}}    & {~=~}    &{0}    \\
\end{array}}$

◼ Remarks:
In [(4) magenta color], we eliminate '2a', by using the information from [(2) magenta color].

3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{x\frac{d^2y}{dx^2}~-~\frac{dy}{dx}~=~0}$ represents a family of parabolas having axis along +ve y-axis.

4. Take any member of the family. Say $\small{y~=~4 x^2~+~3}$
• This member will be a solution of the differential equation that we wrote in step (3) above. Let us check. It can be checked in two steps:

(i) Find the derivatives from the given solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{4 x^2~+~3}    \\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{dy}{dx}}    & {~=~}    &{8x~+~0}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{d^2 y}{dx^2}}    & {~=~}    &{8}    \\
\end{array}}$

(ii) Substitute the above derivatives in the differential equation:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x\frac{d^2y}{dx^2}~-~\frac{dy}{dx}}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{8x~-~8x}    & {~=~}    &{0}    \\
\end{array}}$

Which is true.

• So $\small{y~=~4 x^2~+~3}$ is indeed a solution.

5. A summary of the analysis can be written in three steps:
(i) We were given an algebraic equation. It represents a family of parabolas. It has two arbitrary constants 'a' and 'C'.
(ii) Based on that algebraic equation, we obtained a differential equation. It does not have the arbitrary constants 'a' and 'C'.
(iii) Any member of the family mentioned in (i), is a solution of the differential equation mentioned in (ii).

---

The reader may note the three differences between Example 3 and Example 4.
(i) Family representation
• In (3), the family is represented by:
$\small{y~=~3 x^2~+~\rm{C}}$
• In (4), the family is represented by:
$\small{y~=~a x^2~+~\rm{C}}$

(ii) Coefficient of x²
• In (3), the coefficient of x² is fixed.
• In (4), the coefficient of x² can take different values.

(iii) Elimination of constants
• In (3) we eliminated only one constant 'C'.
• In (4) we eliminated two constants 'a' and 'C'.

---

Example 5
This can be written in 5 steps:

1. Consider the equation $\small{y~=~m x~+~\rm{C}}$.

• We know that, it is the equation of a line.
• By giving different values for 'm' and 'C', we can obtain infinite number of lines which belong to a family. In fact, the equation represents the family of all straight lines in the x-y plane. Some of those lines are shown in the fig.25.10 below:

Fig.25.10

   ♦ The green line has m = 2 and C = 4
   ♦ The red line has m = 1 and C = −2
   ♦ The yellow line has m = −0.2 and C = 1.
so on . . .

2. The above equation contains two arbitrary constants 'm' and 'C'. On many occasions that we encounter in science and engineering, we will want an equation which does not contain arbitrary constants.

• In the present case, let us try to find such an equation. For that, we differentiate the equation as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{m x~+~\rm{C}}    \\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{dy}{dx}}    & {~=~}    &{m}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{d^2y}{dx^2}}    & {~=~}    &{0}    \\
\end{array}}$

(We need the second derivative because, in the first derivative, the constant 'm' is present)

3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{\frac{d^2 y}{dx^2}~=~m}$ represents the family of all straight lines.

4. Take any member of the family. Say $\small{y~=~2 x~+~3}$
• This member will be a solution of the differential equation that we wrote in step (3) above. Let us check. It can be checked in two steps:

(i) Find the derivatives from the given solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{2 x~+~3}    \\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{dy}{dx}}    & {~=~}    &{2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{d^2 y}{dx^2}}    & {~=~}    &{0}    \\
\end{array}}$

(ii) Substitute the above second derivative in the differential equation:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{d^2y}{dx^2}}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{0}    & {~=~}    &{0}    \\
\end{array}}$

Which is true.

• So $\small{y~=~2 x~+~3}$ is indeed a solution.

5. A summary of the analysis can be written in three steps:
(i) We were given an algebraic equation. It represents a family of all straight lines. It has two arbitrary constants 'm' and 'C'.
(ii) Based on that algebraic equation, we obtained a differential equation. It does not have the arbitrary constants 'm' and 'C'.
(iii) Any member of the family mentioned in (i), is a solution of the differential equation mentioned in (ii)

Example 6
This can be written in 5 steps:

1. Consider the equation $\small{y~=~m x}$.

• We know that, it is the equation of a line passing through the origin.
• By giving different values for 'm', we can obtain infinite number of lines which belong to a family. In fact, the equation represents the family of all straight lines lying in the x-y plane and passing through the origin. Some of those lines are shown in the fig.25.11 below:

Fig.25.11

   ♦ The green line has m = 2
   ♦ The red line has m = 1
   ♦ The yellow line has m = −0.2
so on . . .

2. The above equation contains an arbitrary constant 'm'. On many occasions that we encounter in science and engineering, we will want an equation which does not contain arbitrary constants.

• In the present case, let us try to find such an equation. For that, we differentiate the equation as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{m x}    \\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{dy}{dx}}    & {~=~}    &{m}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{y}{x}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x \frac{dy}{dx}}    & {~=~}    &{y}    \\
\end{array}}$

3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{x \frac{dy}{dx}~=~y}$ represents the family of all straight lines passing through the origin.

4. Take any member of the family. Say $\small{y~=~2 x}$
• This member will be a solution of the differential equation that we wrote in step (3) above. Let us check. It can be checked in two steps:

(i) Find the derivatives from the given solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{2 x}    \\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{dy}{dx}}    & {~=~}    &{2}    \\
\end{array}}$

(ii) Substitute the above second derivative in the differential equation:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x \frac{dy}{dx}}    & {~=~}    &{y}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{x(2)}    & {~=~}    &{y}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x(2)}    & {~=~}    &{2x}    \\
\end{array}}$

Which is true.

• So $\small{y~=~2 x}$ is indeed a solution.

5. A summary of the analysis can be written in three steps:
(i) We were given an algebraic equation. It represents a family of all straight lines passing through the origin. It has an arbitrary constant 'm'.
(ii) Based on that algebraic equation, we obtained a differential equation. It does not have the arbitrary constant 'm'.
(iii) Any member of the family mentioned in (i), is a solution of the differential equation mentioned in (ii)

---

After seeing the above 6 examples, we have a basic idea about formation of a differential equation whose general solution is given. 

• We can write four general rules:
(a) The order of the differential equation will be same as the number of arbitrary constants in the given solution.

(b) The given solution needs to be differentiated as many times, as the number of arbitrary constants in that solution.

(c) The derivatives and the original solution, together can be considered as a system of equations. This system can be solved to eliminate the arbitrary constants.

(d) Sometimes it is possible to manipulate the derivatives to eliminate the arbitrary constants.
We did such a manipulation in example 4. In the place of '2a', we wrote: $\small{\frac{dy/dx}{x}}$.

We will now see a solved example.

Solved example 25.16
The equation $\frac{x}{a}~+~\frac{y}{b}~=~1$ gives a family of curves.
Form a differential equation representing the above family of curves, by eliminating arbitrary constants a and b.
Solution:
1. Consider the equation $\frac{x}{a}~+~\frac{y}{b}~=~1$.

• It represents the family of all straight lines with x-intercept 'a' and y-intercept 'b'.

2. To eliminate 'a' and 'b', we differentiate the equation as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{x}{a}~+~\frac{y}{b}}    & {~=~}    &{1}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{1}{a}(1)~+~\frac{1}{b}\left(\frac{dy}{dx} \right)}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{1}{b}\left(\frac{dy}{dx} \right)}    & {~=~}    &{(-1)\frac{1}{a}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\frac{dy}{dx}}    & {~=~}    &{(-1)\frac{b}{a}}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{ \frac{d^2y}{dx^2}}    & {~=~}    &{0}    \\
\end{array}}$

3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{\frac{d^2y}{dx^2}~=~0}$ represents the family of all straight lines with x-intercept 'a' and y-intercept 'b'.
• Note that, in example 4 above, we obtained the same result. The family in example 4, is same as the family in the present solved example. This is because, all straight lines not passing through the origin, will have an x-intercept and a y-intercept.

---

In the next section, we will see a few more solved examples.

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25.2 - Order and Degree of A Differential Equation

In the previous section, we completed a discussion on general and particular solution of differential equations. In this section, we will see order and degree of a differential equation.

• First we will see the various notations used to represent derivatives.

(i) First order derivative can be written as:
${\frac{dy}{dx}~~\text{or}~~y'}$

(ii) Second order derivative can be written as:
${\frac{d^2y}{dx^2}~~\rm{or}~~\it{y\,''}}$  

(iii) Third order derivative can be written as:
${\frac{d^3y}{dx^3}~~\rm{or}~~\it{y\,'''}}$  

(iv) For higher order derivatives, it is inconvenient to use large number of dashes as superscripts. So the nth order derivative can be written as:
${\frac{d^ny}{dx^n}~~\rm{or}~~\it{y_n}}$ 

Order of a differential equation

• Order of a differential equation is defined as the order of the highest order derivative in that given differential equation.
• So the order of a differential equation can be determined in just two steps:
(i) Identify the highest order derivative in the given differential equation.
(ii) The order of that derivative, is the order of the differential equation.

Let us see some examples:
Example 1:
• The given differential equation is: $\small{\frac{dy}{dx}~=~x^2 + 1}$
• There is only one derivative. It's order is 1.
• So the order of the differential equation is 1

Example 2:
• The given differential equation is: $\small{3\frac{d^2y}{dx^2}~+~4\frac{dy}{dx}-2x^2~=~0}$
• There are two derivatives: $\small{\frac{d^2y}{dx^2}}$ and $\small{\frac{dy}{dx}}$
• The highest order derivative is: $\small{\frac{d^2y}{dx^2}}$. It's order is 2.
• So the order of the differential equation is 2

Example 3:
• The given differential equation is: $\small{\frac{d^3y}{dx^3}~+~x^2\left(\frac{d^2y}{dx^2} \right)^3~=~0}$
• There are two derivatives: $\small{\frac{d^3y}{dx^3}}$ and $\small{\frac{d^2y}{dx^2}}$
• The highest order derivative is: $\small{\frac{d^3y}{dx^3}}$. It's order is 3.
• So the order of the differential equation is 3.

---

Degree of a differential equation

• This can be explained in 5 steps:

1. To find the degree of a given differential equation, it is essential that, the differential equation is in the polynomial form.

2. We already know the features of polynomials. Let us recall the main features through some examples:

• ${f(x)=4x^3+5x^2+7}$ is a polynomial function.    

• ${f(x)=4x^3+5x^2+\sin(x)}$ is not a polynomial function.

• ${f(x)=4x^3+5\sqrt{x}+7}$ is not a polynomial function.

3. To make such a comparison for differential equations, assume that, in the place of 'x', we have derivatives. So we can write:

• ${4\left(\frac{dy}{dx} \right)^3~+~5\left(\frac{d^2 y}{dx^2} \right)^2+7}$ is a differential equation in the polynomial form.    

• ${4\left(\frac{d^3y}{dx^3} \right)^2~+~5\left(\frac{dy}{dx} \right)+\sin x}$ is a differential equation in the polynomial form.

• ${4\left(\frac{dy}{dx} \right)^3~+~\sin\left(\frac{dy}{dx} \right)+7}$ is a differential equation. But it is not in the polynomial form.

• ${4\left(\frac{d^3y}{dx^3} \right)^2~+~\sqrt{\frac{dy}{dx}}+7}$ is a differential equation. But it is not in the polynomial form.

4. If the given differential equation is in the polynomial form, we can write the degree in two steps:
(i) Identify the highest order derivative in the given differential equation.
(ii) The power of that derivative, is the degree of the differential equation.

5. If the given differential equation is not in the polynomial form, we say that, degree of that differential equation is not defined.

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Let us see some examples:

Example 1:
• The given differential equation is: $\small{\frac{dy}{dx}~=~x^2 + 1}$
• It is in the polynomial form and hence, degree is defined.
• There is only one derivative.
   ♦ It's order is 1.
   ♦ It's power is 1.
• So for the differential equation:
   ♦ Order is 1.
   ♦ Degree is 1.

Example 2:
• The given differential equation is: $\small{3\frac{d^2y}{dx^2}~+~4\frac{dy}{dx}-2x^2~=~0}$
• It is in the polynomial form and hence, degree is defined.
• There are two derivatives: $\small{\frac{d^2y}{dx^2}}$ and $\small{\frac{dy}{dx}}$
• The highest order derivative is: $\small{\frac{d^2y}{dx^2}}$.
   ♦ It's order is 2.
   ♦ It's power is 1.
• So for the differential equation:
   ♦ Order is 2.
   ♦ Degree is 1.

Example 3:
• The given differential equation is: $\small{\frac{d^3y}{dx^3}~+~x^2\left(\frac{d^2y}{dx^2} \right)^3~=~0}$
• It is in the polynomial form and hence, degree is defined.
• There are two derivatives: $\small{\frac{d^3y}{dx^3}}$ and $\small{\frac{d^2y}{dx^2}}$
• The highest order derivative is: $\small{\frac{d^3y}{dx^3}}$.
   ♦ It's order is 3.
   ♦ It's power is 1.
• So for the differential equation:
   ♦ Order is 3.
   ♦ Degree is 1.

Example 4:
• The given differential equation is: $\small{\frac{dy}{dx}~=~e^x}$
• It is in the polynomial form and hence, degree is defined.
• There is only one derivative.
   ♦ It's order is 1.
   ♦ It's power is 1.
• So for the differential equation:
   ♦ Order is 1.
   ♦ Degree is 1.

Example 5:
• The given differential equation is: $\small{\frac{d^2y}{dx^2}~+~y~=~0}$
• It is in the polynomial form and hence, degree is defined.
• There is only one derivative.
   ♦ It's order is 2.
   ♦ It's power is 1.
• So for the differential equation:
   ♦ Order is 2.
   ♦ Degree is 1.

Example 6:
• The given differential equation is: $\small{\frac{d^3y}{dx^3}~+~x^2\left(\frac{d^2y}{dx^2} \right)^3~=~0}$
• It is in the polynomial form and hence, degree is defined.
• There are two derivatives: $\small{\frac{d^3y}{dx^3}}$ and $\small{\frac{d^2y}{dx^2}}$
• The highest order derivative is: $\small{\frac{d^3y}{dx^3}}$.
   ♦ It's order is 3.
   ♦ It's power is 1.
• So for the differential equation:
   ♦ Order is 3.
   ♦ Degree is 1.

Example 7:
• The given differential equation is: $\small{\frac{d^3y}{dx^3}~+~2\left(\frac{d^2y}{dx^2} \right)^2~-~\frac{dy}{dx}~+~y~=~0}$
• It is in the polynomial form and hence, degree is defined.
• There are three derivatives: $\small{\frac{d^3y}{dx^3},~~\frac{d^2y}{dx^2}}$ and $\small{\frac{dy}{dx}}$
• The highest order derivative is: $\small{\frac{d^3y}{dx^3}}$.
   ♦ It's order is 3.
   ♦ It's power is 1.
• So for the differential equation:
   ♦ Order is 3.
   ♦ Degree is 1.

Example 8:
• The given differential equation is: $\small{\left(\frac{dy}{dx} \right)^2~+~\frac{dy}{dx}~-~\sin^2 y~=~0}$
• It is in the polynomial form and hence, degree is defined.
• There is only one derivative: $\small{\frac{dy}{dx}}$
   ♦ It's order is 1.
   ♦ It's highest power is 2.
• So for the differential equation:
   ♦ Order is 1.
   ♦ Degree is 2.

Example 9:
• The given differential equation is: $\small{\frac{dy}{dx}~+~\sin\left(\frac{dy}{dx} \right)~=~0}$
• It is not in the polynomial form and hence, degree is not defined.
• There is only one derivative: $\small{\frac{dy}{dx}}$
   ♦ It's order is 1.  
• So for the differential equation:
   ♦ Order is 1.

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We can write two important points:
1. We see that, the order of a differential equation is related to the order of a derivative in that differential equation. We know that, order of any derivative is a positive integer.
• So order of a differential equation will be always a positive integer.

2. We see that, degree (if defined) of a differential equation is related to the power of a derivative. When the degree is defined, the differential equation will be in the polynomial form. In the polynomial form, the powers are all positive integers.
• So degree of a differential equation will be always a positive integer.

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Now we will see some solved examples

Solved example 25.14
Find the order of each of the following differential equations. Find also the degree (if defined):

$\small{(i)~~\frac{dy}{dx} - \cos x~=~0}$
$\small{(ii)~~xy \frac{d^2 y}{d x^2} ~+~x \left(\frac{dy}{dx} \right)^2~-~y \frac{dy}{dx}~=~0}$
$\small{(iii)~~y^{'\,'\,'}~+~y^2~+~e^{y'}~=~0}$

Solution:
Part (i):
• The given differential equation is:
$\small{\frac{dy}{dx} - \cos x~=~0}$
• It is in the polynomial form and hence, degree is defined.
• There is only one derivative:
$\small{\frac{dy}{dx}}$
   ♦ It's order is 1.
   ♦ It's power is 1.
• So for the differential equation:
   ♦ Order is 1.
   ♦ Degree is 1.

Part (ii):
• The given differential equation is:
$\small{xy \frac{d^2 y}{d x^2} ~+~x \left(\frac{dy}{dx} \right)^2~-~y \frac{dy}{dx}~=~0}$
• It is in the polynomial form and hence, degree is defined.
• There are two derivatives: $\small{\frac{d^2y}{dx^2}}$ and $\small{\frac{dy}{dx}}$
• The highest order derivative is: $\small{\frac{d^2y}{dx^2}}$.
   ♦ It's order is 2.
   ♦ It's power is 1.
• So for the differential equation:
   ♦ Order is 2.
   ♦ Degree is 1.

Part (iii):
• The given differential equation is:
$\small{y^{'\,'\,'}~+~y^2~+~e^{y'}~=~0}$
• It is not in the polynomial form and hence, degree is not defined.
• There is only one derivative: $\small{y^{'\,'\,'}}$
   ♦ It's order is 3.
• So for the differential equation:
   ♦ Order is 3.

Solved example 25.15
Find the order of each of the following differential equations. Find also the degree (if defined):

$\small{(i)~~\left(\frac{d^2y}{dx^2} \right)^2 + \cos \left(\frac{dy}{dx} \right)~=~0}$

$\small{(ii)~~y^{'\,'\,'} ~+~2y^{'\,'}~+~y' ~=~0}$

Solution:
Part (i):
• The given differential equation is:
 $\small{\left(\frac{d^2y}{dx^2} \right)^2 + \cos \left(\frac{dy}{dx} \right)~=~0}$
• It is not in the polynomial form and hence, degree is not defined.
• There are two derivatives: $\small{\frac{d^2y}{dx^2}~~\rm{and}~~\frac{dy}{dx}}$
• The highest order derivative is: $\small{\frac{d^2y}{dx^2}}$.
   ♦ It's order is 2.  
• So for the differential equation:
   ♦ Order is 2.

Part (ii):
• The given differential equation is:
$\small{y^{'\,'\,'} ~+~2y^{'\,'}~+~y' ~=~0}$
• It is in the polynomial form and hence, degree is defined.
• There are three derivatives: $\small{y^{'\,'\,'},~y^{'\,'}~~\rm{and}~~y'}$
• The highest order derivative is: $\small{y^{'\,'\,'}}$.
   ♦ It's order is 3.
   ♦ It's power is 1.
• So for the differential equation:
   ♦ Order is 3.
   ♦ Degree is 1.

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The links below gives a few more solved examples:

Exercise 25.1

Exercise 25.2

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In the next section, we will see formation of differential equations when general solution is given.

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25.1 - Solved Examples on General and Particular Solutions of a Differential Equation

In the previous section, we saw how to check whether a function is the general solution or particular solution of a given differential equation. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved example 25.7
(a) Solve the differential equation $\small{\frac{dy}{dt}~=~-t}$
Given that:
$\small{y=1}$ when $\small{t = 0}$.

(b) Solve the differential equation $\small{\frac{dy}{dt}~=~-t}$
Given that:
$\small{y=-1}$ when $\small{t = 0}$.

(c) Draw both results in the same graph

Solution:
Part (a):
1. We have: $\small{\frac{dy}{dt}~=~-t}$
• Integrating both sides, we get:
$\small{y + \rm{C}_1~=~\frac{(-1)t^2}{2} + \rm{C}_2}$
$\small{\Rightarrow y~=~\frac{(-1)t^2}{2} + \rm{C}_2 - \rm{C}_1}$
$\small{\Rightarrow y~=~\frac{(-1)t^2}{2} + \rm{C}}$
• This is the general solution.

2. Given that:
$\small{y=1}$ when $\small{t = 0}$.
• Substituting in the general solution, we get:
$\small{1~=~\frac{(-1)(0)^2}{2} + \rm{C}}$
$\small{\Rightarrow 1~=~0 + \rm{C}}$
$\small{\Rightarrow 1~=~ \rm{C}}$

3. So the particular solution is:
$\small{y~=~\frac{(-1)t^2}{2} + 1}$

Part (b):
1. From part (a), we have:
$\small{y~=~\frac{(-1)t^2}{2} + \rm{C}}$
• This is the general solution.

2. Given that:
$\small{y=-1}$ when $\small{t = 0}$.
• Substituting in the general solution, we get:
$\small{-1~=~\frac{(-1)(0)^2}{2} + \rm{C}}$
$\small{\Rightarrow -1~=~0 + \rm{C}}$
$\small{\Rightarrow -1~=~ \rm{C}}$

3. So the particular solution is:
$\small{y~=~\frac{(-1)t^2}{2} - 1}$

Part (c):
The graphs are shown in fig.25.4 below:

Fig.25.4

• The green curve represents the particular solution for part (a)
• The red curve represents the particular solution for part (b)

Solved example 25.8
(a) Solve the differential equation $\small{\frac{dy}{dt}~=~2}$
Given that:
$\small{y=1}$ when $\small{t = 0}$.

(b) Solve the differential equation $\small{\frac{dy}{dt}~=~2}$
Given that:
$\small{y=-1}$ when $\small{t = 0}$.

(c) Draw both results in the same graph

Solution:
Part (a):
1. We have: $\small{\frac{dy}{dt}~=~2}$
• Integrating both sides, we get:
$\small{y + \rm{C}_1~=~2t + \rm{C}_2}$
$\small{\Rightarrow y~=~2t + \rm{C}_2 - \rm{C}_1}$
$\small{\Rightarrow y~=~2t + \rm{C}}$
• This is the general solution.

2. Given that:
$\small{y=1}$ when $\small{t = 0}$.
• Substituting in the general solution, we get:
$\small{1~=~2(0) + \rm{C}}$
$\small{\Rightarrow 1~=~0 + \rm{C}}$
$\small{\Rightarrow 1~=~ \rm{C}}$

3. So the particular solution is:
$\small{y~=~2t + 1}$

Part (b):
1. From part (a), we have:
$\small{y~=~2t + \rm{C}}$
• This is the general solution.

2. Given that:
$\small{y=-1}$ when $\small{t = 0}$.
• Substituting in the general solution, we get:
$\small{-1~=~2(0) + \rm{C}}$
$\small{\Rightarrow -1~=~0 + \rm{C}}$
$\small{\Rightarrow -1~=~ \rm{C}}$

3. So the particular solution is:
$\small{y~=~2t - 1}$

Part (c):
The graphs are shown in fig.25.5 below:

Fig.25.5

• The green curve represents the particular solution for part (a)
• The red curve represents the particular solution for part (b)

Solved example 25.9
(a) Solve the differential equation $\small{\frac{dv}{dt}~=~4t}$
Given that:
$\small{v=10}$ when $\small{t = 0}$.

(b) At what time does v increase to 100 or drop to 1?

Solution:
Part (a):
1. We have: $\small{\frac{dv}{dt}~=~4t}$
• Integrating both sides, we get:
$\small{v + \rm{C}_1~=~\frac{4t^2}{2} + \rm{C}_2}$
$\small{\Rightarrow v~=~2 t^2 + \rm{C}_2 - \rm{C}_1}$
$\small{\Rightarrow v~=~2t^2 + \rm{C}}$
• This is the general solution.

2. Given that:
$\small{v=10}$ when $\small{t = 0}$.
• Substituting in the general solution, we get:
$\small{10~=~2(0)^2 + \rm{C}}$
$\small{\Rightarrow 10~=~0 + \rm{C}}$
$\small{\Rightarrow 10~=~ \rm{C}}$

3. So the particular solution is:
$\small{v~=~2t^2 + 10}$

Part (b):
1. From part (a), we have:
$\small{v~=~2t^2 + 10}$
• This is the particular solution for the differential equation.

2. Substitute v = 100 in the particular solution:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{100}    & {~=~}    &{2 t^2~+~10}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{2t^2}    & {~=~}    &{90}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{t^2}    & {~=~}    &{45}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{t}    & {~=~}    &{3 \sqrt 5}    \\
\end{array}}$                           

(We discard the −ve root because, time cannot be −ve.

3. We can write:
The velocity increase to 100 at the instant when the reading in the stop-watch is $\small{3 \sqrt 5}$ seconds.

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An analysis of the above solved example 25.9, will give us greater insight into the basics of differential equations. The analysis can be written in 4 steps:

1. We are given the differential equation: $\small{\frac{dv}{dt}~=~4t}$
• On the L.H.S, we have the derivative of velocity w.r.t time. So that derivative is the rate of change of velocity w.r.t time. In other words, it is the acceleration.
• We can write: Acceleration of the given object is given by (4t)

2. If we integrate $\small{\frac{dv}{dt}}$, we will get the velocity v.
• So if we integrate (4t), we will get the velocity.
• That means, $\small{2t^2 + \rm{C}}$ is the expression for velocity.
• This can be considered as the expression for velocity, for any object moving with an acceleration of (4t)

3. Our particular object has an initial value for velocity: When t = 0, v = 10.
• Using this initial value, we obtained the useful result:
At any instant t, our particular object will be traveling with a velocity of $\small{2t^2 + 10}$
• Note that, $\small{2t^2 + 10}$ is valid only for our particular object. For another object, even if acceleration is (4t), the initial value may be different and so the expression may be different.

4. We are asked to find the time when the velocity becomes 100 or 1.
• For our particular object, the velocity can never become 1. The reason can be written in 3 steps:
(i) The acceleration is (4t), which is +ve. The initial velocity is 10.
(ii) A velocity of one, is less than a velocity of ten.
(iii) A lesser velocity can be achieved only if the object undergoes deceleration.   
• In other words, a lesser velocity can be achieved only if the object undergoes −ve acceleration. Our particular object has a +ve acceleration.

---

Solved example 25.10
Verify that the function $\small{y~=~e^{-3x}}$ is a solution of the differential equation $\small{\frac{d^2y}{dx^2}~+~\frac{dy}{dx}~-~6y~=~0}$

Solution:
1. Find the derivatives from the given solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{e^{-3x}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{{\frac{dy}{dx}}}    & {~=~}    &{(-3)e^{-3x}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{{\frac{d^2y}{dx^2}}}    & {~=~}    &{(-3)(-3)e^{-3x}~=~9 e^{-3x}}    \\
\end{array}}$                           

2. Substitute the above derivatives in the given differential equation:

• The given differential equation is: $\small{\frac{d^2y}{dx^2}~+~\frac{dy}{dx}~-~6y~=~0}$

Substituting, we get: $\small{9 e^{-3x}~+~(-3) e^{-3x}~-~6e^{-3x}~=~0}$
Which is true.

• So $\small{y = e^{-3x}}$ is indeed a solution.

Solved example 25.11
Verify that the function $\small{y~=~a \cos x~+~b \sin x}$, where a, b ∈ R is a solution of the differential equation $\small{\frac{d^2y}{dx^2}~+~y~=~0}$

Solution:
1. Find the derivatives from the given solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{a \cos x~+~b \sin x}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{{\frac{dy}{dx}}}    & {~=~}    &{-a \sin x~+~b \cos x}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{{\frac{d^2y}{dx^2}}}    & {~=~}    &{-a \cos x~-~b \sin x}    \\
\end{array}}$                           

2. Substitute the above derivatives in the given differential equation:

• The given differential equation is: $\small{\frac{d^2y}{dx^2}~+~y~=~0}$

Substituting, we get: $\small{\left[-a \cos x~-~b \sin x \right]~+~\left[a \cos x~+~b \sin x \right]~=~0}$

$\small{\Rightarrow a(\cos x~-~\cos x)~+~b(\sin x~-~\sin x)~=~0}$
Which is true.

• So $\small{y = a \cos x~+~b \sin x}$ is indeed a solution.

Solved example 25.12
Verify that the implicit function $\small{y~-~\cos y~=~x}$, is a solution of the differential equation $\small{(y \sin y ~+~\cos y~+~x)\frac{dy}{dx}~=~y}$

Solution:
1. We already know the method to find $\small{\frac{dy}{dx}}$ when we are given an implicit function. (see section 21.8) 

• So in out present case, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y~-~\cos y}    & {~=~}    &{x}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{{\frac{dy}{dx}~-~\left((-\sin y)\frac{dy}{dx} \right)}}    & {~=~}    &{1}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{{\frac{dy}{dx}~+~\left(\sin y \right)\frac{dy}{dx}}}    & {~=~}    &{1}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{{\left(1~+~\sin y \right)\frac{dy}{dx}}}    & {~=~}    &{1}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{{\frac{dy}{dx}}}    & {~=~}    &{\frac{1}{1~+~\sin y}}    \\
\end{array}}$                           

2. Substitute the above derivative in the given differential equation:

• The given differential equation is: $\small{(y \sin y ~+~\cos y~+~x)\frac{dy}{dx}~=~y}$

Substituting, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{(y \sin y ~+~\cos y~+~x)\frac{1}{1~+~\sin y}}    & {~=~}    &{y}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{{(y \sin y ~+~\cos y~+~y~-~\cos y)\frac{1}{1~+~\sin y}}}    & {~=~}    &{y}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{{(y \sin y ~+~y)\frac{1}{1~+~\sin y}}}    & {~=~}    &{y}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{{y(\sin y ~+~1)\frac{1}{1~+~\sin y}}}    & {~=~}    &{y}    \\
\end{array}}$

Which is true.

• So the implicit function  $\small{y-\cos y~=~x}$ is indeed a solution.

Solved example 25.13
Verify that the implicit function $\small{x~+~y~=~\tan^{-1} y}$, is a solution of the differential equation $\small{y^2\frac{dy}{dx}~+~y^2~+~1~=~0}$

Solution:
1. We already know the method to find $\small{\frac{dy}{dx}}$ when we are given an implicit function. (see section 21.8) 

• So in out present case, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x~+~y}    & {~=~}    &{\tan^{-1}y}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{{1~+~\frac{dy}{dx}}}    & {~=~}    &{\frac{1}{1 + y^2}\frac{dy}{dx}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{1}{1 + y^2}\frac{dy}{dx}~-~\frac{dy}{dx}}    & {~=~}    &{1}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\frac{dy}{dx}\left(\frac{1}{1 + y^2}~-~1 \right)}    & {~=~}    &{1}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{\frac{dy}{dx}\left(\frac{1-1 - y^2}{1 + y^2} \right)}    & {~=~}    &{1}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{\frac{dy}{dx}\left(\frac{- y^2}{1 + y^2} \right)}    & {~=~}    &{1}    \\
{~\color{magenta}    7    }    &{{\Rightarrow}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{-1- y^2}{y^2}}    \\
\end{array}}$                           

2. Substitute the above derivative in the given differential equation:

• The given differential equation is: $\small{y^2\frac{dy}{dx}~+~y^2~+~1~=~0}$

Substituting, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y^2\left(\frac{-1- y^2}{y^2} \right)~+~y^2~+~1}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{-1- y^2~+~y^2~+~1}    & {~=~}    &{0}    \\
\end{array}}$

Which is true.

• So the implicit function  $\bf{x~+~y~=~\tan^{-1} y}$ is indeed a solution.

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In the next section, we will see order and degree of differential equations.

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Chapter 25 - Differential Equations

In the previous section, we completed a discussion on application of integrals. In this chapter, we will see differential equations.

Some basic details can be written in 8 steps:

1. Consider a car moving with a constant acceleration 'a'. We know that, the distance traveled by the car is given by the equation: $\small{s = ut + \frac{1}{2}at^2}$
Where,
s = Distance traveled in time t
u = the initial velocity (a constant)
a = acceleration with which the car is traveling (a constant)

• So distance depends on t. In other words, distance is a function of t.
   ♦ s is the dependent variable.
   ♦ t is the independent variable.
• We can write: $\small{s = f(t) = \text{u}t + \frac{1}{2}\text{a}t^2}$

2. We know that, velocity is the derivative of distance w.r.t time. So we can write:

$\small{v = f'(t) = \frac{ds}{dt}=\text{u} + \frac{1}{2}\text{a}(2t)}$

$\small{\Rightarrow v = f'(t) = \frac{ds}{dt}=\text{u} + \text{a}t}$

$\small{\Rightarrow v =\text{u} + \text{a}t}$

3. We also know that, acceleration  is the derivative of velocity w.r.t time. So we can write:

$\small{\frac{dv}{dt} = f''(t) = \frac{d^2s}{dt^2}=0 + \text{a}}$

$\small{\Rightarrow \frac{dv}{dt} = f''(t) = \frac{d^2s}{dt^2}=\text{a}}$  

$\small{\Rightarrow \frac{dv}{dt}=\text{a}}$  

4. Let us now think in a reverse manner. If we are given the acceleration, can we find the velocity?
• Note that, the car is moving with a constant acceleration 'a'. So the velocity is continuously changing.
• Then the question is:
If we are given the acceleration 'a', can we write an expression for the velocity?
(If we can write such an expression, we will be able to find the velocity at any instant t)
• Let us try to write such an expression:
In step (3), we wrote: $\small{\frac{dv}{dt}=\text{a}}$

Integrating both sides, we get:
$\small{v + \rm{C}_1=\text{a}t + \rm{C}_2}$  

$\small{\Rightarrow v =\text{a}t + \rm{C}_2 - \rm{C}_1}$  

$\small{\Rightarrow v =\text{a}t + \rm{C}}$

5. The expression that we obtained in the above step (4), is applicable to any object moving with a constant acceleration 'a'.
• It is called the general solution of the differential equation $\small{\frac{dv}{dt}=\text{a}}$.

6. To put the above general equation to practical use, we need to find the value of the constant 'C'. For that, we must have some additional information.
• Let, at the instant when the stop-watch is turned on, the car be moving at a velocity of 9 m/s
• At the time when the stop-watch is turned on, time t = 0.
• Substituting these values in the general solution, we get:
$\small{9 =\text{a}(0) + \rm{C}}$
$\small{\Rightarrow 9 =\rm{C}}$
• So for the car, we can write the expression for velocity:
$\small{v =\text{a}t + 9}$
• This is called particular solution of the differential equation $\small{\frac{dv}{dt}=\text{a}}$.
(Note that, this particular solution is same as our familiar equation $\small{v =\text{u} + \text{a}t}$  where u is the initial velocity)
• We were able to write the particular solution by using the fact that:
Initially, when t = 0, velocity is 9 m/s

7. The following graph in fig.25.1 will help us to understand the difference between general solution and particular solution. It is assumed that, the constant a = 3/4 ms⁻².

Fig.25.1

• The topmost red line represents $\small{v =\frac{3}{4} t + 14}$, where the constant C = 14  
• The second red line from top, represents $\small{v =\frac{3}{4} t + 11}$, where the constant C = 11
• In this way, we can draw infinite number of parallel red lines.
• But we are interested in the green line which represents the unique particular solution, where C = 9
• Note that, the graph of the particular solution, passes through (0,9).
• (0,9) is an ordered pair.
   ♦ The x-coordinate (abscissa) gives time    
   ♦ The y-coordinate (ordinate) gives velocity
It is a velocity-time graph.

8. From the above steps, we learned six new items:
(i) We saw the differential equation $\small{\frac{dv}{dt}=\text{a}}$
(ii) In step (4), we got the general solution of that differential equation. We got:
$\small{v =\text{a}t + \rm{C}}$
(iii) In step (6), we got the particular solution of that differential equation. We got:
$\small{v =\text{a}t + 9}$
(iv) Differential equation is a relation between:
   ♦ Derivative of the dependent variable v
   ♦ and
   ♦ The independent variable t
(v) Solution (general or particular) of the differential equation is a relation between:
   ♦ Dependent variable v
   ♦ and
   ♦ The independent variable t
(vi) In the graph of step(7), we saw the difference between general solution and particular solution.

---

Let us see another example. It can be written in 7 steps:

1. Consider the same car that we saw in the example above. It is moving with a constant acceleration 'a'.

2. We saw that,

$\small{v = f'(t) = \frac{ds}{dt}=\text{u} + \text{a}t}$

3. Let us now think in a reverse manner. If we are given the velocity, can we find the distance traveled?
• Note that, the car is moving with a constant acceleration 'a'. So the velocity is continuously changing. Consequently, the distance traveled in unit time is also continuously changing.
• Then the question is:
If we are given the expression for velocity, can we write an expression for the distance traveled?
(If we can write such an expression, we will be able to find the distance traveled any instant t)
• Let us try to write such an expression:
In step (2), we wrote: $\small{\frac{ds}{dt}=\text{u} + \text{a}t}$ Integrating both sides, we get:
$\small{s + \rm{C}_1=\text{u}t + \text{a} \frac{t^2}{2}+ \rm{C}_2}$  
$\small{\Rightarrow s =\text{u}t + \frac{1}{2} \text{a} t^2 + \rm{C}_2 - \rm{C}_1}$  
$\small{\Rightarrow s =\text{u}t + \frac{1}{2} \text{a} t^2 + \rm{C}}$  
4. The expression that we obtained in the above step (3), is applicable to any object moving with a constant acceleration 'a' and initial velocity u.
• It is called the general solution of the differential equation $\small{\frac{ds}{dt}=\text{u} + \text{a}t}$

5. To put the above general equation to practical use, we need to find the value of the constant 'C'. For that, we must have some additional information.
• At the instant when the stop-watch is turned on, time t = 0.
• In a duration of zero seconds, the distance 's' traveled will be zero.
• Substituting these values in the general solution, we get:
$\small{0 =\text{u}(0) + \frac{1}{2} \text{a} (0)^2 + \rm{C}}$  
$\small{\Rightarrow 0 = 0 + 0 + \rm{C}}$  
$\small{\Rightarrow 0 =\rm{C}}$
• So for the car, we can write the expression for distance:
$\small{s =\text{u}t + \frac{1}{2} \text{a} t^2 + 0}$  
$\small{\Rightarrow s =\text{u}t + \frac{1}{2} \text{a} t^2}$  
• This is called particular solution of the differential equation $\small{\frac{ds}{dt}=\text{u} + \text{a}t}$.
(Note that, this particular solution is same as our familiar equation $\small{s =\text{u} t +\frac{1}{2} \text{a} t^2}$  where u is the initial velocity)
• We were able to write the particular solution by using the fact that:
Initially, when t = 0, distance traveled is zero.
6. The following graph in fig.25.2 will help us to understand the difference between general solution and particular solution. It is assumed that,
   ♦ the constant a = 3/4 ms⁻².
   ♦ the constant u = 9 ms⁻¹.

Fig.25.2

• The topmost red curve represents $\small{s =\text{u}t + \frac{1}{2} \text{a} t^2 + 19}$  , where the constant C = 19  
• The second red curve from top, represents $\small{s =\text{u}t + \frac{1}{2} \text{a} t^2 + 15}$  , where the constant C = 15  
• In this way, we can draw infinite number of red curves.
• But we are interested in the green curve which represents the unique particular solution, where C = 0
• Note that, the graph of the particular solution, passes through (0,0).
• (0,0) is an ordered pair.
   ♦ The x-coordinate (abscissa) gives time    
   ♦ The y-coordinate (ordinate) gives distance
It is a distance-time graph.

7. From the above steps, we learned four new items:
(i) We saw a differential equation $\small{\frac{ds}{dt}=\text{u} + \text{a}t}$
(ii) In step (3), we got the general solution of that differential equation. We got:
$\small{s =\text{u}t + \frac{1}{2} \text{a} t^2 + \rm{C}}$  
(iii) In step (5), we got the particular solution of that differential equation. We got:
$\small{s =\text{u}t + \frac{1}{2} \text{a} t^2}$ 
(iv) Differential equation is a relation between:
   ♦ Derivative of the dependent variable s
   ♦ and
   ♦ The independent variable t
(v) Solution (general or particular) of the differential equation is a relation between:
   ♦ Dependent variable s
   ♦ and
   ♦ The independent variable t
• In the graph of step(6), we saw the difference between general solution and particular solution.

---

Now we will see an important feature of the general solution. It can be written in two steps:
1. A solution must satisfy the given equation. For example:
If someone says that x = 8 is a solution of the equation x² − 6x −16 = 0, we can cross check by substituting x = 8.
• We get: 8² − 6(8) − 16 = 0
⇒ 64 − 48 − 16 = 0
⇒ 16 − 16 = 0, which is true.
• So x = 8 is indeed a solution of the equation x² − 6x −16 = 0

2. In our present case, we saw how the general solution of a differential equation can be obtained. After obtaining the general solution, we must cross check. (In some cases, we will be asked to cross check a given solution) The checking can be done in 4 steps.

◼ Consider the first example.
(i) We obtained $\small{v =\text{a}t + \rm{C}}$ as the general solution for the differential equation $\small{\frac{dv}{dt}=\text{a}}$.
(ii) Differentiating the general solution, we get: $\small{\frac{dv}{dt} = \text {a}}$
(iii) Substituting this in the given differential equation, we get: $\small{\text{a}=\text{a}}$, which is true.
• So $\small{v =\text{a}t + \rm{C}}$ is indeed the general solution of the given differential equation.
(iv) If particular solution is available, then it must also be cross checked.
• In the first example, the particular solution is: $\small{v =\text{a}t + 9}$.
• We have the additional information that, when t = 0, velocity is 9. In other words, (0,9) is a point in the solution.
• Substituting this point in the particular solution, we get: $\small{9 =\text{a}(0) + 9}$    
$\small{\Rightarrow 9 = 0 + 9}$, which is true.    
• So $\small{v =\text{a}t + 9}$ is indeed the particular solution of the given differential equation.

◼ Consider the second example.
(i) We obtained $\small{s =\text{u}t + \frac{1}{2} \text{a} t^2 + \rm{C}}$ as the general solution for the differential equation $\small{\frac{ds}{dt}=\text{u} + \text{a}t}$.
(ii) Differentiating the general solution, we get: $\small{\frac{ds}{dt} = \text {u} + \frac{1}{2} \text{a} (2t)  + 0}$
$\small{\Rightarrow \frac{ds}{dt} = \text {u} + \text{a}t}$
(iii) Substituting this in the given differential equation, we get:
$\small{\text {u} + \text{a}t=\text{u} + \text{a}t}$, which is true.
• So $\small{s =\text{u}t + \frac{1}{2} \text{a} t^2 + \rm{C}}$ is indeed the general solution of the given differential equation.
(iv) If particular solution is available, then it must also be cross checked.
• In the second example, the particular solution is: $\small{s =\text{u}t + \frac{1}{2} \text{a} t^2}$.
• We have the additional information that, when t = 0, distance is zero. In other words, (0,0) is a point in the solution.
Substituting this point in the particular solution, we get: $\small{0 =\text{u}(0) + \frac{1}{2} \text{a} (0)^2}$    
$\small{\Rightarrow 0 = 0 + 0}$, which is true.    
• So $\small{s =\text{u}t + \frac{1}{2} \text{a} t^2}$ is indeed the particular solution of the given differential equation.

---

Now we will see some solved examples:

Solved example 25.1
Check whether the function $\small{y = \frac{4x^3}{3} + \rm{C}}$ is the general solution of the differential equation $\small{\frac{dy}{dx} = 4x^2}$

If $\small{y=-30}$ when $\small{x = -3}$, then find the particular solution.
Solution:
Part I: Checking whether the given function is the general solution.

1. Find the derivative $\small{\frac{dy}{dx}}$ from the given general solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{\frac{4x^3}{3} + \rm{C}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{{\frac{dy}{dx}}}    & {~=~}    &{\frac{4}{3} \left(3x^2 \right)+0}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{4x^2}    \\
\end{array}}$                           

2. Substitute the above derivative in the given differential equation:

• The given differential equation is: $\small{\frac{dy}{dx} = 4x^2}$

Substituting, we get: $\small{4x^2 = 4x^2}$
Which is true.

• So $\small{y = \frac{4x^3}{3} + \rm{C}}$ is indeed the general solution.

Part II: Finding the particular solution.

1. The general solution contains the constant C. We must find the value of C.
• Given that, $\small{y=-30}$ when $\small{x = -3}$
• Substituting these in the general solution, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{\frac{4x^3}{3} + \rm{C}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{-30}    & {~=~}    &{\frac{4(-3)^3}{3} + \rm{C}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{-30}    & {~=~}    &{4(-1)3^2 + \rm{C}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{-30 + 36}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{6}    & {~=~}    &{\rm{C}}    \\
\end{array}}$                           

2. So the particular solution is:

$\small{y = \frac{4x^3}{3} + 6}$

Solved example 25.2
Check whether the function $\small{y = \frac{3x^4}{4} + \rm{C}}$ is the general solution of the differential equation $\small{\frac{dy}{dx} = 3x^3}$

If $\small{y=\frac{19}{4}}$ when $\small{x = 1}$, then find the particular solution.
Solution:
Part I: Checking whether the given function is the general solution.

1. Find the derivative $\small{\frac{dy}{dx}}$ from the given general solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{\frac{3x^4}{4} + \rm{C}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{{\frac{dy}{dx}}}    & {~=~}    &{\frac{3}{4} \left(4x^3 \right)+0}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{3x^3}    \\
\end{array}}$                           

2. Substitute the above derivative in the given differential equation:

• The given differential equation is: $\small{\frac{dy}{dx} = 3x^3}$

Substituting, we get: $\small{3x^3 = 3x^3}$
Which is true.

• So $\small{y = \frac{3x^4}{4} + \rm{C}}$ is indeed the general solution.

Part II: Finding the particular solution.

1. The general solution contains the constant C. We must find the value of C.
• Given that, $\small{y=\frac{19}{4}}$ when $\small{x = 1}$
• Substituting these in the general solution, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{\frac{3x^4}{4} + \rm{C}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{19}{4}}    & {~=~}    &{\frac{3(1)^4}{4} + \rm{C}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{19}{4}}    & {~=~}    &{\frac{3}{4} + \rm{C}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\frac{16}{4}}    & {~=~}    &{\rm{C}}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{4}    & {~=~}    &{\rm{C}}    \\
\end{array}}$                           

2. So the particular solution is:

$\small{y = \frac{3x^4}{4} + 4}$

Solved example 25.3
Check whether the function $\small{y = \rm{C}\,e^{x^2}}$ is the general solution of the differential equation $\small{\frac{dy}{dx} = 2xy}$

If $\small{y=\frac{1}{2}}$ when $\small{x = 0}$, then find the particular solution.
Solution:
Part I: Checking whether the given function is the general solution.

1. Find the derivative $\small{\frac{dy}{dx}}$ from the given general solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{\rm{C}\,e^{x^2}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{{\frac{dy}{dx}}}    & {~=~}    &{\rm{C}\,e^{x^2}(2x)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{2\rm{C}\,xe^{x^2}}    \\
\end{array}}$                           

2. Substitute the above derivative and y in the given differential equation:

• The given differential equation is: $\small{\frac{dy}{dx} = 3xy}$

Substituting, we get: $\small{2\rm{C}\,xe^{x^2} = 2x(\rm{C}\,e^{x^2})}$

$\small{\Rightarrow 2\,xe^{x^2} = 2x(e^{x^2})}$
Which is true.

• So $\small{y = \rm{C}\,e^{x^2}}$ is indeed the general solution.

Part II: Finding the particular solution.

1. The general solution contains the constant C. We must find the value of C.
• Given that, $\small{y=\frac{1}{2}}$ when $\small{x = 0}$
• Substituting these in the general solution, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{\rm{C}\,e^{x^2}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{1}{2}}    & {~=~}    &{\rm{C}\,e^{(0)^2}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{1}{2}}    & {~=~}    &{\rm{C}\,e^{0}}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\frac{1}{2}}    & {~=~}    &{\rm{C}(1)}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{\frac{1}{2}}    & {~=~}    &{\rm{C}}    \\
\end{array}}$                           

2. So the particular solution is:

$\small{y = \frac{1}{2} e^{x^2}}$

3. The graph is shown in fig.25.3 below:

Fig.25.3

• The top most red curve represents the solution when C = 2
• The second red curve from top represents the solution when C = 1
• The bottom most red curve represents the solution when C = 1/3
• In this manner, infinite number of red curves can be drawn. But there is only one green curve, which represents the particular solution. For that green curve, C = 1/2.

Solved example 25.4
Check whether the function $\small{y = \rm{C}\,e^{-1/x}}$ is the general solution of the differential equation $\small{\frac{dy}{dx} x^2 = y}$

If $\small{y=\frac{2}{e}}$ when $\small{x = 1}$, then find the particular solution.
Solution:
Part I: Checking whether the given function is the general solution.

1. Find the derivative $\small{\frac{dy}{dx}}$ from the given general solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{\rm{C}\,e^{-1/x}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{{\frac{dy}{dx}}}    & {~=~}    &{\rm{C}\,e^{-1/x}\left(\frac{1}{x^2} \right)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{\rm{C}\,e^{-1/x}}{x^2} }    \\
\end{array}}$                           

2. Substitute the above derivative and y in the given differential equation:

• The given differential equation is: $\small{\frac{dy}{dx} x^2 = y}$

Substituting, we get: $\small{\left(\frac{\rm{C}\,e^{-1/x}}{x^2} \right) x^2 = \rm{C}\,e^{-1/x}}$

Which is true.

• So $\small{y = \rm{C}\,e^{-1/x}}$ is indeed the general solution.

Part II: Finding the particular solution.

1. The general solution contains the constant C. We must find the value of C.
• Given that, $\small{y=\frac{2}{e}}$ when $\small{x = 1}$
• Substituting these in the general solution, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{\rm{C}\,e^{-1/x}}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{2}{e}}    & {~=~}    &{\rm{C}\,e^{-1/1}}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{2}{e}}    & {~=~}    &{\rm{C}\left(\frac{1}{e} \right)}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\frac{2}{e}}    & {~=~}    &{\frac{\rm{C}}{e}}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{2}    & {~=~}    &{\rm{C}}    \\
\end{array}}$                           

2. So the particular solution is:

$\small{y = 2 e^{-1/x}}$

Solved example 25.5
Check whether the function $\small{u = \sin^{-1}\left(e^{\rm{C} + t} \right)}$ is the general solution of the differential equation $\small{\frac{du}{dt} = \tan u}$

If $\small{u=\frac{\pi}{2}}$ when $\small{t = 1}$, then find the particular solution.
Solution:
Part I: Checking whether the given function is the general solution.

1. Find the derivative $\small{\frac{du}{dt}}$ from the given general solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{u}    & {~=~}    &{\sin^{-1}\left(e^{\rm{C} + t} \right)}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{{\frac{du}{dt}}}    & {~=~}    &{\frac{e^{\rm{C} + t}}{\sqrt{1 - \left(e^{\rm{C} + t} \right)^2 }}}    \\
\end{array}}$                           

2. Substitute the above derivative and u in the given differential equation:

• The given differential equation is: $\small{\frac{du}{dt} = \tan u}$

Substituting, we get: $\small{\frac{e^{\rm{C} + t}}{\sqrt{1 - \left(e^{\rm{C} + t} \right)^2 }} = \tan\left[\sin^{-1}\left(e^{\rm{C} + t} \right) \right]}$

3. So our next task is to find: $\small{ \tan\left[\sin^{-1}\left(e^{\rm{C} + t} \right) \right]}$

Let $\small{y = \sin^{-1}\left(e^{\rm{C} + t} \right)}$. Then $\small{\sin y = e^{\rm{C} + t}}$

we want tan y.

• Since $\small{\sin y = e^{\rm{C} + t}}$, we can write:

$\small{\cos y = \sqrt{1~-~\left(e^{\rm{C} + t} \right)^2}}$

• Then $\small{\tan y~=~\frac{e^{\rm{C} + t}}{\sqrt{1~-~\left(e^{\rm{C} + t} \right)^2}}}$

4. So from step (2), we get:

$\small{\frac{e^{\rm{C} + t}}{\sqrt{1 - \left(e^{\rm{C} + t} \right)^2 }} = \tan\left[\sin^{-1}\left(e^{\rm{C} + t} \right) \right]~=~\frac{e^{\rm{C} + t}}{\sqrt{1~-~\left(e^{\rm{C} + t} \right)^2}}}$

Which is true.

• So $\small{u = \sin^{-1}\left(e^{\rm{C} + t} \right)}$ is indeed the general solution.

Part II: Finding the particular solution.

1. The general solution contains the constant C. We must find the value of C.
• Given that, $\small{u=\frac{\pi}{2}}$ when $\small{t = 1}$
• Substituting these in the general solution, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{u}    & {~=~}    &{\sin^{-1}\left(e^{\rm{C}+t} \right)}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{\pi}{2}}    & {~=~}    &{\sin^{-1}\left(e^{\rm{C}+1} \right)}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\sin \left(\frac{\pi}{2} \right)}    & {~=~}    &{e^{\rm{C}+1} }    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{1}    & {~=~}    &{e(e^{\rm{C}})}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{e^{\rm{C}}}    & {~=~}    &{\frac{1}{e}~=~e^{-1}}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{\rm{C}}    & {~=~}    &{-1}    \\
\end{array}}$                           

2. So the particular solution is:

$\small{u = \sin^{-1}\left(e^{-1+t} \right)}$

Solved example 25.6
Check whether the function $\small{x = \rm{C}~-~\frac{1}{8} \sin(2t)~-~\frac{1}{32} \sin(4t)}$ is the general solution of the differential equation $\small{8 \frac{dx}{dt} = -2 \cos (2t)~-~\cos(4t)}$

If $\small{x=\pi}$ when $\small{t = \pi}$, then find the particular solution.
Solution:
Part I: Checking whether the given function is the general solution.

1. Find the derivative $\small{\frac{dx}{dt}}$ from the given general solution:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x}    & {~=~}    &{\rm{C}~-~\frac{1}{8} \sin(2t)~-~\frac{1}{32} \sin(4t)}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{{\frac{dx}{dt}}}    & {~=~}    &{0~-~\frac{1}{8} (2)\cos(2t)~-~\frac{1}{32} (4)\cos(4t)}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{{\frac{dx}{dt}}}    & {~=~}    &{-\frac{1}{4}\cos(2t)~-~\frac{1}{8} \cos(4t)}    \\
\end{array}}$                           

2. Substitute the above derivative and x in the given differential equation:

• The given differential equation is: $\small{8 \frac{dx}{dt} = -2 \cos (2t)~-~\cos(4t)}$

Substituting, we get:

$\small{8 \left[-\frac{1}{4}\cos(2t)~-~\frac{1}{8} \cos(4t) \right] = -2 \cos (2t)~-~\cos(4t)}$

$\small{\Rightarrow  \left[-2 \cos(2t)~-~ \cos(4t) \right] = -2 \cos (2t)~-~\cos(4t)}$

Which is true.

• So $\small{x = \rm{C}~-~\frac{1}{8} \sin(2t)~-~\frac{1}{32} \sin(4t)}$ is indeed the general solution.

Part II: Finding the particular solution.

1. The general solution contains the constant C. We must find the value of C.
• Given that, $\small{x=\pi}$ when $\small{t = \pi}$
• Substituting these in the general solution, we get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x}    & {~=~}    &{\rm{C}~-~\frac{1}{8} \sin(2t)~-~\frac{1}{32} \sin(4t)}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\pi}    & {~=~}    &{\rm{C}~-~\frac{1}{8} \sin(2\pi)~-~\frac{1}{32} \sin(4 \pi)}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\pi}    & {~=~}    &{\rm{C}~-~\frac{1}{8} (0)~-~\frac{1}{32} (0)}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{\pi}    & {~=~}    &{\rm{C}~-~0}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{\rm{C}}    & {~=~}    &{\pi}    \\
\end{array}}$                           

2. So the particular solution is:

$\small{x ~=~ \pi~-~\frac{1}{8} \sin(2t)~-~\frac{1}{32} \sin(4t)}$

---

In the next section, we will see a few more solved examples.

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