25.4 - Solved Examples Related To Formation of Differential Equation

In the previous section, we saw formation of a differential equation whose general solution is given. We saw a solved example also. In this section, we will see a few more solved examples.

Solved example 25.17
The equation $y^2~=~a(b^2 - x^2)$ gives a family of curves.
Form a differential equation representing the above family of curves, by eliminating arbitrary constants a and b.
Solution:
1. To eliminate 'a' and 'b', we differentiate the equation as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y^2}    & {~=~}    &{a(b^2 - x^2)}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{2y\,y'}    & {~=~}    &{0~-~2x\,a}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{y\,y'}    & {~=~}    &{-ax}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{y\,y''~+~y'\,y'}    & {~=~}    &{-a}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{y\,y''~+~(y')^2}    & {~=~}    &{\frac{y\,y'}{x}}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{xy\,y''~+~x\,(y')^2~-~y\,y'}    & {~=~}    &{0}    \\
\end{array}}$

◼ Remarks:
In [(5) magenta color], we use the result from [(3) magenta color] to eliminate 'a'

2. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{xy\,y''~+~x\,(y')^2~-~y\,y'~=~0}$ represents the given family of curves.

Solved example 25.18
The equation $y~=~a \sin(x+b)$ gives a family of curves.
Form a differential equation representing the above family of curves, by eliminating arbitrary constants a and b.
Solution:
1. To eliminate 'a' and 'b', we differentiate the equation as shown below:

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

◼ Remarks:
In [(4) magenta color], we use the original equation from [(1) magenta color] to eliminate 'a' and 'b'

2. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{y''~+~y~=~0}$ represents the given family of curves.

Solved example 25.19
Form the differential equation representing the family of ellipses having foci on the x-axis and center at the origin.
Solution:
1. We know that, the family mentioned in the question is given by the equation: $\small{\frac{x^2}{a^2}~+~\frac{y^2}{b^2}~=~1}$ (see section 11.4)

A few members of the family are shown in the fig.25.12 below:

Fig.25.12

For the green ellipse, a = 5 and b = 3
For the red ellipse, a = 3 and b = 2
For the yellow ellipse, a = 7 and b = 4

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

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{x^2}{a^2}~+~\frac{y^2}{b^2}}    & {~=~}    &{1}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{2x}{a^2}~+~\frac{2y\,y'}{b^2}}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{x}{a^2}~+~\frac{y\,y'}{b^2}}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x b^2~+~y\,y'\,a^2}    & {~=~}    &{0}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{ b^2~+~a^2\left[y\,y''~+~(y')^2 \right]}    & {~=~}    &{0}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{ \frac{(-1)\,y\,y'\,a^2}{x}~+~a^2\left[y\,y''~+~(y')^2 \right]}    & {~=~}    &{0}    \\
{~\color{magenta}    7    }    &{{\Rightarrow}}    &{ \frac{(-1)\,y\,y'}{x}~+~y\,y''~+~(y')^2}    & {~=~}    &{0}    \\
{~\color{magenta}    8    }    &{{\Rightarrow}}    &{xy\,y''~+~x(y')^2~-~y\,y'}    & {~=~}    &{0}    \\
\end{array}}$

◼ Remarks:
In [(6) magenta color], we use the result from [(4) magenta color] to eliminate $\small{b^2}$

3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{xy\,y''~+~x(y')^2~-~y\,y'~=~0}$ represents the given family of curves.

Solved example 25.20
Form the differential equation representing the family of ellipses having foci on the y-axis and center at the origin.
Solution:
1. We know that, the family mentioned in the question is given by the equation: $\small{\frac{x^2}{b^2}~+~\frac{y^2}{a^2}~=~1}$

A few members of the family are shown in the fig.25.13 below:

Fig.25.13

For the green ellipse, a = 3 and b = 5
For the red ellipse, a = 2 and b = 3
For the yellow ellipse, a = 4 and b = 7

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

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{x^2}{b^2}~+~\frac{y^2}{a^2}}    & {~=~}    &{1}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{2x}{b^2}~+~\frac{2y\,y'}{a^2}}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{x}{b^2}~+~\frac{y\,y'}{a^2}}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x a^2~+~y\,y'\,b^2}    & {~=~}    &{0}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{ a^2~+~b^2\left[y\,y''~+~(y')^2 \right]}    & {~=~}    &{0}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{ \frac{(-1)\,y\,y'\,b^2}{x}~+~b^2\left[y\,y''~+~(y')^2 \right]}    & {~=~}    &{0}    \\
{~\color{magenta}    7    }    &{{\Rightarrow}}    &{ \frac{(-1)\,y\,y'}{x}~+~y\,y''~+~(y')^2}    & {~=~}    &{0}    \\
{~\color{magenta}    8    }    &{{\Rightarrow}}    &{xy\,y''~+~x(y')^2~-~y\,y'}    & {~=~}    &{0}    \\
\end{array}}$

◼ Remarks:
In [(6) magenta color], we use the result from [(4) magenta color] to eliminate $\small{a^2}$

3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{xy\,y''~+~x(y')^2~-~y\,y'~=~0}$ represents the given family of curves.

4. This is the same result as in the previous solved example 25.19. The reason can be written in two steps:
(i) The algebraic equation for the two types of ellipses are basically the same. The only difference is that, the arbitrary constants 'a' and 'b' are interchanged.
(ii) In the differential equation, there is no role for the arbitrary constants. So we get the same result. 

Solved example 25.21
Form the differential equation representing the family of circles touching the x-axis at the origin.
Solution:
1. We know that, the circle with center at (a,b) and radius r is given by the equation: $\small{(x-a)^2~+~(y-b)^2~=~r^2}$
The family mentioned in the question will be similar to the one shown in fig.25.14 below:

Fig.25.14

• Based on the fig., we can write:
'a' will be zero and 'b' will be equal to 'r'.

• So the equation of the family becomes:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x^2~+~(y-r)^2}    & {~=~}    &{r^2}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{x^2~+~y^2~-~2yr~+~r^2}    & {~=~}    &{r^2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x^2~+~y^2~-~2yr}    & {~=~}    &{0}    \\
\end{array}}$

• In the above fig.25.14,
   ♦ r = 1 for the green circle
   ♦ r = 3 for the red circle
   ♦ r = 2 for the yellow circle

2. To eliminate 'r', we differentiate the equation as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x^2~+~y^2~-~2yr}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{2x~+~2y\,y'~-~2r\,y'}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x~+~y\,y'~-~r\,y'}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x~+~(y~-~r)\,y'}    & {~=~}    &{0}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{x~+~\left[y~-~\left(\frac{x^2~+~y^2}{2y} \right)\right]\,y'}    & {~=~}    &{0}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{x~+~\left[\frac{2y^2~-~x^2~-~y^2}{2y}\right]\,y'}    & {~=~}    &{0}    \\
{~\color{magenta}    7    }    &{{\Rightarrow}}    &{x~+~\left[\frac{y^2~-~x^2}{2y}\right]\,y'}    & {~=~}    &{0}    \\
{~\color{magenta}    8    }    &{{\Rightarrow}}    &{\left[\frac{x^2~-~y^2}{2y}\right]\,y'}    & {~=~}    &{x}    \\
{~\color{magenta}    9    }    &{{\Rightarrow}}    &{y'}    & {~=~}    &{\frac{2xy}{x^2~-~y^2}}    \\
\end{array}}$

◼ Remarks:
In [(5) magenta color], we use the original equation from [(1) magenta color] to eliminate $\small{r}$

3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{y'~=~\frac{2xy}{x^2~-~y^2}}$ represents the given family of curves.

Solved example 25.22
Form the differential equation representing the family of circles touching the y-axis at the origin.
Solution:
1. We know that, the circle with center at (a,b) and radius r is given by the equation: $\small{(x-a)^2~+~(y-b)^2~=~r^2}$
The family mentioned in the question will be similar to the one shown in fig.25.15 below:

Fig.25.15

• Based on the fig., we can write:
'b' will be zero and 'a' will be equal to 'r'.

• So the equation of the family becomes:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{(x-r)^2~+~y^2}    & {~=~}    &{r^2}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{x^2~-~2xr~+~r^2~+~y^2}    & {~=~}    &{r^2}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x^2~+~y^2~-~2xr}    & {~=~}    &{0}    \\
\end{array}}$

• In the above fig.25.15,
   ♦ r = 1 for the green circle
   ♦ r = 3 for the red circle
   ♦ r = 2 for the yellow circle

2. To eliminate 'r', we differentiate the equation as shown below:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x^2~+~y^2~-~2xr}    & {~=~}    &{0}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{2x~+~2y\,y'~-~2r}    & {~=~}    &{0}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x~+~y\,y'~-~r}    & {~=~}    &{0}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x~+~y\,y'~-~\left(\frac{x^2~+~y^2}{2x} \right)}    & {~=~}    &{0}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{\left[x~-~\left(\frac{x^2~+~y^2}{2x} \right) \right]~+~y\,y'}    & {~=~}    &{0}    \\
{~\color{magenta}    6    }    &{{\Rightarrow}}    &{\left[\frac{2x^2~-~x^2~-~y^2}{2x} \right]~+~y\,y'}    & {~=~}    &{0}    \\
{~\color{magenta}    7    }    &{{\Rightarrow}}    &{\left[\frac{x^2~-~y^2}{2x} \right]~+~y\,y'}    & {~=~}    &{0}    \\
{~\color{magenta}    8    }    &{{\Rightarrow}}    &{y\,y'}    & {~=~}    &{\left[\frac{y^2~-~x^2}{2x} \right]}    \\
{~\color{magenta}    9    }    &{{\Rightarrow}}    &{y'}    & {~=~}    &{\frac{y^2~-~x^2}{2xy}}    \\
\end{array}}$

◼ Remarks:
In [(4) magenta color], we use the original equation from [(1) magenta color] to eliminate $\small{r}$

3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{y'~=~\frac{y^2~-~x^2}{2xy}}$ represents the given family of curves.

Solved example 25.23
Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of x-axis.
Solution:
1. We know that, the family mentioned in the question is given by the equation: $\small{y^2~=~4ax}$

A few members of the family are shown in the fig.25.16 below:

Fig.25.16

For the green parabola, a = 2
For the red parabola, a = 4
For the yellow parabola, a = 3

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

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y^2}    & {~=~}    &{4ax}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{2y\,y'}    & {~=~}    &{4a}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{y\,y'}    & {~=~}    &{2a}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{y\,y'}    & {~=~}    &{2\left(\frac{y^2}{4x} \right)}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}    &{y'}    & {~=~}    &{\frac{y}{2x}}    \\
\end{array}}$

◼ Remarks:
In [(4) magenta color], we use the original equation from [(1) magenta color] to eliminate $\small{a}$

3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{y'~=~\frac{y}{2x}}$ represents the given family of curves.

Solved example 25.24
Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of y-axis.
Solution:
1. We know that, the family mentioned in the question is given by the equation: $\small{x^2~=~4ay}$

A few members of the family are shown in the fig.25.17 below:

Fig.25.17

For the green parabola, a = 2
For the red parabola, a = 4
For the yellow parabola, a = 3

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

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x^2}    & {~=~}    &{4ay}    \\
{~\color{magenta}    2    }    &{{\Rightarrow}}    &{2x}    & {~=~}    &{4a\,y'}    \\
{~\color{magenta}    3    }    &{{\Rightarrow}}    &{x}    & {~=~}    &{2a\,y'}    \\
{~\color{magenta}    4    }    &{{\Rightarrow}}    &{x}    & {~=~}    &{y'\left(\frac{x^2}{2y} \right)}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}        &{1}    & {~=~}    &{y'\left(\frac{x}{2y} \right)}    \\
{~\color{magenta}    5    }    &{{\Rightarrow}}        &{y'}    & {~=~}    &{\frac{2y}{x}}    \\
\end{array}}$

◼ Remarks:
In [(4) magenta color], we use the original equation from [(1) magenta color] to eliminate $\small{a}$

3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{y'~=~\frac{2y}{x}}$ represents the given family of curves.

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The link below gives a few more solved examples:

Exercise 9.3

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In the next section, we will see solution of differential equations.

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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.

---

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.

---

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.

---

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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