Formation of Differential Equation and Solutions of a Differential Equation

What is a Differential Equation?

A differential equation is an equation involving one or more terms and the derivatives of one dependent variable with respect to the other independent variable.

Differential equation: dy/dx = f(x)

Where “x” is an independent variable and “y” is a dependent variable

Example of differential equation: $x \frac{d y}{d x}+2 y=0$
The above-written equation involves variables as well as the derivative of the dependent variable $\mathrm{y}$ with respect to the independent variable $\mathrm{x}$. Therefore, it is a differential equation.
The following relations are some of the examples of differential equations:
(i) $\frac{d y}{d x}=\sin 2 x+\cos x$
(ii) $\mathrm{k} \frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^2\right]^{3 / 2}$

Formation of Differential Equation

We have studied the general formula of the parabola which is y² = 4ax

This equation represents a family of parabolas with an arbitrary constant. With different values of a, we get a different parabola in this family.

To form its D.E, let us first differentiate it w.r.t. x,

$
2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=4 \mathrm{a} \quad \text { or } \quad \frac{\mathrm{y}}{2} \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{a}
$

Putting this value of 'a' in the original equation of the parabola

$
\begin{aligned}
& y^2=4 \cdot \frac{y}{2} \frac{d y}{d x} \cdot x \\
& 2 x \frac{d y}{d x}=y
\end{aligned}
$
This is the Differential Equation for a family of parabolas y² = 4ax.

Note that there is one arbitrary constant in the original equation, and the order of the D.E. formed is also 1. This is an important result: if an equation contains n arbitrary constants, then its D.E. will have an order equal to n.

If we are given a relation between the variables x, y with n arbitrary constants C₁, C₂, .... C_n, then to form its D.E., we differentiate the given relation n times in succession with respect to x, we have n + 1 equations altogether. Now using these we eliminate the n arbitrary constants. The result is a differential equation of the nth order.

Consider the equation of the family of ellipses with a and b as arbitrary constants

This is the D.E. that represents the given family of ellipses

Consider the equation of the family of ellipses with $a$ and $b$ as arbitrary constants

$
\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1
$

Differentiate Eq (4) w.r.t. 'x ${ }^{\prime}$

$
\frac{2 \mathrm{x}}{\mathrm{a}^2}+\frac{2 \mathrm{y}}{\mathrm{b}^2} \cdot\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=0 \quad \text { or } \quad-\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{\mathrm{yy}^{\prime}}{\mathrm{x}}
$

Differentiate above equation w.r.t. ' $x$ '

$
\begin{gathered}
0=\frac{\mathrm{x}\left(\mathrm{y}^{\prime 2}+\mathrm{yy}^{\prime \prime}\right)-\mathrm{yy}^{\prime}}{\mathrm{x}^2} \\
\mathrm{yy}^{\prime}=\mathrm{xyy}^{\prime \prime}+\mathrm{x}\left(\mathrm{y}^{\prime}\right)^2
\end{gathered}
$

Illustration :

Differential equation of the equation $y=(a+b) e^x+e^{x+c}$ is Here, the number of the arbitrary constant is $3: a, b$, and $c$. But we can club arbitrary constants together

$
y=\left(a+b+e^c\right) e^x
$

which is of the form, $\mathrm{y}=\mathrm{Ae}^{\mathrm{x}} \quad$ where, $\mathrm{A}=\left(\mathrm{a}+\mathrm{b}+\mathrm{e}^{\mathrm{c}}\right)$ hence, corresponding Differential equation will be of order 1

$
\begin{aligned}
& \Rightarrow \frac{d y}{d x}=A e^x \\
& \Rightarrow \frac{d y}{d x}=y
\end{aligned}
$

Solution of Differential Equations

Solutions can be classified according to different types of equations and given conditions. There are different types of solutions for differential equations based on their types or structure differential equations. General solutions and particular solutions are two types of differential equation solutions.

If the given D.E. is

$
\begin{aligned}
& 2 x \frac{d y}{d x}=y \\
& 2 \frac{d y}{y}=\frac{d x}{x} \\
& 2 \int \frac{d y}{y}=\int \frac{d x}{x} \\
& 2 \ln (y)=\ln (x)+c
\end{aligned}
$

As $c$ is a constant of integration, it can take any real value. We can write it as $\ln (4 a)$ as well, where a is now the arbitrary constant.

$
\begin{aligned}
& \ln \left(y^2\right)=\ln (x)+\ln (4 a) \\
& y^2=4 a x
\end{aligned}
$
This is the general solution of the differential equation, which represents the family of the parabola

If we put some real value of a in this equation, then we get a particular solution of the given differential equation. For example, if we put a = 2, then y² = 8x is a particular solution of this equation.

Hence, the solution of the differential equation is a relation between the variables of the equation that satisfy the D.E. and does not contain any derivatives or any arbitrary constants.

A general solution of a differential equation is a relation between the variables (not involving the derivatives) which contains the same number of the arbitrary constants as the order of the differential equation.

A particular solution of the differential equation is obtained from the general solution by assigning particular values to the arbitrary constant in the general solution.

To solve the first-order differential equation of first degree, some standard forms are available to get the general solution. They are:

• Variable separable method
• Reducible into the variable separable method
• Linear differential equation
• Reducible into a linear differential equation
• Exact differential equations
• Linear differential equations with constant coefficients

Note: If arbitrary constants appear in addition, subtraction, multiplication, or division, then we can club them to reduce into one new arbitrary constant. Hence, the differential equation corresponding to a family of curves will have the order the same as the number of essential arbitrary constants (number of arbitrary constants in the modified form) in the curve equation.

Recommended Video Based on Formation of Differential Equations and its Solution

Solved Examples Based On Formation of Differential Equations and its Solution

Example 1: The differential equation for the family of curves $x^2+y^2-2 a y=0$, where $a$ is an arbitrary constant is.
(1) $\left(x^2-y^2\right) y^{\prime}=2 x y$
(2) $2\left(x^2+y^2\right) y^{\prime}=x y$
(3) $2\left(x^2-y^2\right) y^{\prime}=x y$
(4) $\left(x^2+y^2\right) y^{\prime}=2 x y$

Solution:

$
\begin{aligned}
& x^2+y^2-2 a y=0 \\
& 2 x+2 y \frac{d y}{d x}-2 a \cdot \frac{d y}{d x}=0 \\
& \Rightarrow x+y \frac{d y}{d x}=a \frac{d y}{d x} \\
& \Rightarrow x+y \frac{d y}{d x}=\frac{x^2+y^2}{2 y} \cdot \frac{d y}{d x} \\
& \Rightarrow 2 x y+2 y^2 \frac{d y}{d x}=x^2 \frac{d y}{d x}+y^2 \frac{d y}{d x} \\
& \Rightarrow\left(x^2-y^2\right) \frac{d y}{d x}=2 x y \\
& \Rightarrow\left(x^2-y^2\right) y^{\prime}=2 x y
\end{aligned}
$
Hence, the answer is option (1).

Example 2:The differential equation which represents the family of curves $y=c_1 e^{c_2 x}$, where $c_1$ and $c_2$ are arbitrary constants, is
(1) $y^{\prime \prime}=y^{\prime} y$
(2) $y y^{\prime \prime}=y^{\prime}$
(3) $y y^{\prime \prime}=\left(y^{\prime}\right)^2$
(4) $y^{\prime}=y^2$

Solution: $\square$
As we learned in
Formation of Differential Equations -
A differential equation can be derived from its equation by the process of differentiation and another algebraical process of elimination

$
\begin{aligned}
& y=C_1 \cdot e^{C_2 x} \\
& y^{\prime}=C_1 \cdot e^{C_2 x} \times C_2 \Rightarrow C_1 C_2 \cdot e^{C_2 x}=C_2 y \\
& y^{\prime \prime}=C_2 \cdot y^{\prime} \\
& \Rightarrow \frac{y^{\prime}}{y^{\prime \prime}}=\frac{y}{y^{\prime}} \\
& \Rightarrow y y^{\prime \prime}=\left(y^{\prime}\right)^2
\end{aligned}
$

Hence, the answer is the option (3).

Example 3: If the differential equation representing the family of all circles touching the $x$ - axis at the origin is $\left(x^2-y^2\right) \frac{d y}{d x}=g(x) y$, then $g(x)$ equals :
(1) $\frac{1}{2} x$
(2) $2 x^2$
(3) $2 x$
(4) $\frac{1}{2} x^2$

Solution:
As we learned in
Formation of Differential Equations -
A differential equation can be derived from its equation by the process of differentiation and another algebraical process of elimination

$
\begin{aligned}
& \Rightarrow 2 x+2(y-a) \frac{d y}{d x}=0 \\
& \therefore x+(y-a) \frac{d y}{d x}=0 \\
& \therefore \frac{d y}{d x}=\frac{x}{(y-a)} \quad \text { or } \quad a=\frac{x+y \frac{d y}{d y}}{d y / d x}
\end{aligned}
$

Put a in (i)

$
\begin{aligned}
& x^2+\frac{x^2}{(d y / d x)^2}=\left(\frac{x+y \frac{d y}{d x}}{(d y / d x)}\right)^2 \\
& \Rightarrow\left(x^2-y^2\right) \frac{d y}{d x}=2 x y \\
& \therefore g(x)=2 x
\end{aligned}
$
Hence, the answer is option (3).

Example 4: The differential equation whose solution is $A x^2+B y^2=1$, where $A$ and $B$ are arbitrary constant, is of
(1) second-order and second-degree
(2) first-order and second degree
(3) first-order and first-degree
(4) second-order and first-degree

Solution:
After forming the differential equation, its order and degree can be determined.

$
A x^2+B y^2=1
$$

Differentiating, we get:

$
\begin{aligned}
& \Rightarrow 2 A x+2 B y \frac{d y}{d x}=0 \\
& \Rightarrow A x+B y \frac{d y}{d x}=0
\end{aligned}
$$

Differentiating, we get:

$
\begin{aligned}
& \Rightarrow A+B\left[\frac{d y}{d x} \cdot \frac{d y}{d x}+y \cdot \frac{d^2 y}{d x^2}\right]=0 \\
& \Rightarrow A+B\left[\left(\frac{d y}{d x}\right)^2+y \frac{d^2 y}{d x^2}\right]=0
\end{aligned}
$
By eliminating A and B, we get:

$x y y^{\prime \prime}+x\left(y^{\prime}\right)^2-y y^{\prime}=0$

Here, Order = 2 and Degree = 1.

Hence, the answer is option (4).

Summary

The formation of differential equations from a given set of conditions or family of functions involves differentiating and eliminating arbitrary constants. Solutions to differential equations provide functions that satisfy the equations and can be general, particular, or singular. Various methods, including separation of variables, integrating factors, and numerical techniques, are used to solve these equations, making differential equations powerful tools for modeling and analyzing dynamic systems.