Chapter 9: Differential Equations

An equation involving derivatives (or differentials) of the dependent variable with respect to independent variable (or variables) is termed as a differential equation.

Order and Degree of a Differential Equation

• Order: The order of a differential equation is the order of the highest order derivative occurring in the differential equation.
• Degree: The degree of a differential equation is the highest power (positive integral index) of the highest order derivative occurring in it, after it has been expressed as a polynomial of derivatives.

Solutions of Differential Equations

A function which satisfies the given differential equation is called its solution.

• General Solution: The solution which contains arbitrary constants is called the general solution (primitive) of the differential equation.
• Particular Solution: The solution obtained from the general solution by giving particular values to the arbitrary constants is called a particular solution of the differential equation.

Methods of finding the solution of First Order First Degree Differential Equations:

1. Differential Equations with Variables Separable: If the equation $\frac{dy}{dx} = f(x, y)$ can be written in the form $g(y)dy = h(x)dx$, integrating both sides gives the solution: $\int g(y)dy = \int h(x)dx + C$.

2. Homogeneous Differential Equations: An equation $\frac{dy}{dx} = F(x, y)$ is said to be homogeneous if $F(x, y)$ is a homogeneous function of degree zero. To solve it, we make the substitution $y = vx$. Then $\frac{dy}{dx} = v + x\frac{dv}{dx}$ and the equation gets reduced to the separable form.

3. Linear Differential Equations: Equations of the form $\frac{dy}{dx} + Py = Q$, where $P$ and $Q$ are functions of $x$ or constants. Integrating Factor (I.F.) = $e^{\int P dx}$. The solution is given by: $y \times \text{I.F.} = \int (Q \times \text{I.F.}) dx + C$.

Competency-Based Questions

1. [CBSE 2021] Find the specific solution of the differential equation $\cos(x) \frac{dy}{dx} + y \sin x = 1$, given that $y = 1$ when $x = 0$.

Solution:

$\cos x \frac{dy}{dx} + y \sin x = 1$

Divide by $\cos x$:

$\frac{dy}{dx} + y \tan x = \sec x$.

This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$.

Here, $P = \tan x$, $Q = \sec x$.

Integrating Factor (I.F.) $= e^{\int P dx} = e^{\int \tan x dx} = e^{\log|\sec x|} = \sec x$.

The general solution is:

$y \times \text{I.F.} = \int (Q \cdot \text{I.F.}) dx + C$

$y \sec x = \int (\sec x \cdot \sec x) dx + C$

$y \sec x = \int \sec^2 x dx + C$

$y \sec x = \tan x + C$.

Now, to find the specific solution, use the initial conditions: $y = 1$ when $x = 0$.

$(1) \sec(0) = \tan(0) + C$

$1 \times 1 = 0 + C \implies C = 1$.

Therefore, the specific solution is $y \sec x = \tan x + 1$ or $y = \sin x + \cos x$.

2. [Sample Paper 2022] Solve the differential equation $\frac{dy}{dx} = e^{x+y}$.

Solution:

$\frac{dy}{dx} = e^{x+y} = e^x \cdot e^y$

This is a separable variables equation.

$\frac{dy}{e^y} = e^x dx$

$e^{-y} dy = e^x dx$

Integrating both sides:

$\int e^{-y} dy = \int e^x dx$

$-e^{-y} = e^x + C_1$

$e^x + e^{-y} = -C_1$

Let $C = -C_1$, an arbitrary constant.

$e^x + e^{-y} = C$ is the general solution.

3. [CBSE 2019] Show that the differential equation $(x - y) \frac{dy}{dx} = x + 2y$ is homogeneous and solve it.

Solution:

$\frac{dy}{dx} = \frac{x + 2y}{x - y} = F(x, y)$

Let $x \to \lambda x$, $y \to \lambda y$: $F(\lambda x, \lambda y) = \frac{\lambda x + 2\lambda y}{\lambda x - \lambda y} = \frac{\lambda(x + 2y)}{\lambda(x - y)} = \lambda^0 F(x, y)$.

So, it is a homogeneous differential equation.

To solve, put $y = vx$. Then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.

Substituting into the equation:

$v + x \frac{dv}{dx} = \frac{x + 2vx}{x - vx} = \frac{x(1 + 2v)}{x(1 - v)} = \frac{1 + 2v}{1 - v}$

$x \frac{dv}{dx} = \frac{1 + 2v}{1 - v} - v = \frac{1 + 2v - v(1 - v)}{1 - v} = \frac{1 + 2v - v + v^2}{1 - v}$

$x \frac{dv}{dx} = \frac{v^2 + v + 1}{1 - v}$

Separating variables:

$\frac{1 - v}{v^2 + v + 1} dv = \frac{dx}{x}$

Multiply by 2 for the numerator to become the derivative of denominator:

$\frac{2 - 2v}{v^2 + v + 1} dv = \frac{2}{x} dx$

$\frac{3 - (2v + 1)}{v^2 + v + 1} dv = \frac{2}{x} dx$

$\int \frac{3}{v^2 + v + 1} dv - \int \frac{2v + 1}{v^2 + v + 1} dv = \int \frac{2}{x} dx$

$\int \frac{3}{(v + 1/2)^2 + (\sqrt{3}/2)^2} dv - \log|v^2 + v + 1| = 2 \log|x| + C$

$3 \cdot \frac{1}{\sqrt{3}/2} \tan^{-1}\left(\frac{v + 1/2}{\sqrt{3}/2}\right) = \log|v^2 + v + 1| + \log x^2 + C$

$2\sqrt{3} \tan^{-1}\left(\frac{2v + 1}{\sqrt{3}}\right) = \log|x^2(v^2 + v + 1)| + C$

Substitute $v = y/x$:

$2\sqrt{3} \tan^{-1}\left(\frac{2y/x + 1}{\sqrt{3}}\right) = \log|x^2(y^2/x^2 + y/x + 1)| + C$

$2\sqrt{3} \tan^{-1}\left(\frac{2y + x}{\sqrt{3}x}\right) = \log|y^2 + xy + x^2| + C$.