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:
(ii) Coefficient of x2
$\small{y~=~a x^2~+~\rm{C}}$
•
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.
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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