Question

Solve the following differential equation
$\left( {{x}^{2}}-1 \right)\dfrac{dy}{dx}+2xy=\dfrac{2}{{{x}^{2}}-1}$.

Answer 1

Hint: Given a non-homogeneous equation. Try to make it a linear differential equation. And then find an integrating factor, which we can multiply to the both sides of the differential equation so that the Left Hand Side becomes in the form $d\left( \phi \left( x \right) \right)$ .So, that when we integrate on the sides, we can easily solve the integration and hence get a solution.

Complete step by step answer:
We are given a non-homogeneous differential equation.
First, we will try to convert it into the general form of differential equation
By dividing the whole equation with $\left( {{x}^{2}}-1 \right)$
Consider the given differential equation,
$\left( {{x}^{2}}-1 \right)\dfrac{dy}{dx}+2xy=\dfrac{2}{{{x}^{2}}-1}$
On dividing by $\left( {{x}^{2}}-1 \right)$ on both the sides of the equation, we get,
$\dfrac{dy}{dx}+\dfrac{2x}{{{x}^{2}}-1}y=\dfrac{2}{{{\left( {{x}^{2}}-1 \right)}^{2}}}$
Now, it is looking like the general form of first order linear differential equation.
We know that for solving the first order linear differential equation, we first find the integrating factor,
GENERAL FORM OF LINEAR DIFFERENTIAL EQUATION.
An integrating factor is a function by which an ordinary differential equation can be multiplied in order to make it integrable. For example, a linear first-order ordinary differential equation of type
$\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$
Where P and Q are the continuous function, can be made integrable by letting $v\left( x \right)$be a function
$v\left( x \right)=\int_{{}}^{{}}{p\left( x \right)dx}$
Then ${{e}^{v\left( x \right)}}$ would be the integrating factor therefore,
It’s INTEGRATING FACTOR IS
${{e}^{\int{P\left( x \right)dx}}}$
So, the integrating factor for given differential equation is after comparing the general form with the given equation,
${{e}^{\int{\dfrac{2x}{{{x}^{2}}-1}dx}}}$
Let, ${{x}^{2}}-1=t$
On differentiating both the sides,
$2xdx=dt$
So, now the integrating factor becomes,
$\begin{align}
  & {{e}^{\int{\dfrac{1}{t}dt}}} \\
 & ={{e}^{\ln t}} \\
 & =t \\
 & ={{x}^{2}}-1 \\
\end{align}$
Multiplying the integrating factor to the differential equation, then we will get,
$\left( {{x}^{2}}-1 \right)\dfrac{dy}{dx}+2xy=\dfrac{2}{\left( {{x}^{2}}-1 \right)}$
Hence,
$d\left( \left( {{x}^{2}}-1 \right)y \right)=\dfrac{2}{{{x}^{2}}-1}$
Now, integrating both the sides, we get
$\left( {{x}^{2}}-1 \right)y=\int{\dfrac{2}{{{x}^{2}}-1}dx}+C$
We know the formula of integration,
$\int{\dfrac{1}{{{x}^{2}}-{{a}^{2}}}dx}=\dfrac{1}{2a}\ln \left| \dfrac{x-a}{x+a} \right|+C$
Applying this formula in the last step, we get,
$\left( {{x}^{2}}-1 \right)y=2\dfrac{1}{2}\ln \left| \dfrac{x-1}{x+1} \right|+C$

Hence, the solution of the differential equation $\left( {{x}^{2}}-1 \right)\dfrac{dy}{dx}+2xy=\dfrac{2}{{{x}^{2}}-1}$ is
 $\left( {{x}^{2}}-1 \right)y=2\dfrac{1}{2}\ln \left| \dfrac{x-1}{x+1} \right|+C$

Note: There is an alternate method to use here or we can say the method of inspection that we can use here without calculating the integrating factor. Consider the given differential equation $\left( {{x}^{2}}-1 \right)\dfrac{dy}{dx}+2xy=\dfrac{2}{{{x}^{2}}-1}$ , Notice the Left hand side of the differential equation, it is the differentiation of x$y\left( {{x}^{2}}-1 \right)$ with respect to $''x''$ , so we can directly get into the step $d\left( \left( {{x}^{2}}-1 \right)y \right)=\dfrac{2}{{{x}^{2}}-1}$ and proceed as same.