Question

The integrating factor of the differential equation $\left(1+x^2\right)\frac{dy}{dx}+y=e^{{\tan }^{-1}x}$ is

• A. ${\tan }^{-1}x$
• B. $1+x^2$
• C. $e^{{\tan }^{-1}x}$
• D. ${\log }_e\left(1+x^2\right)$

Answer 1

The correct option is C

$e^{{\tan }^{-1}x}$

Explanation for the correct answer:

Step 1: Compare given differential equation with general form of linear differential equation

Given differential equation, $\left(1+x^2\right)\frac{dy}{dx}+y=e^{{\tan }^{-1}x}$.

$\Rightarrow \frac{dy}{dx}+\frac{1}{\left(1+x^2\right)}y=\frac{e^{{\tan }^{-1}x}}{\left(1+x^2\right)}$.

Compare the differential equation with the general form of the linear differential equation $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{+}\mathit{P}\mathit{y}\mathbf{=}\mathit{Q}$.

Here, $P$ and $Q$ are functions of $x$.

Thus, $P=\frac{1}{1+x^2}$.

Step 2: Substitute the value of $P$ in formula for integrating factor

So, the integrating factor of the given differential equation can be provided by, $R=e^{\int Pdx}$.

$\Rightarrow R=e^{\int \left(\frac{1}{1+x^2}\right)dx}$.

Step 3: Use substitution method to complete integration

Let us assume that, $x=\tan z$.

Differentiate both sides of the equation.

$dx={\sec }^{2}zdz$.

Hence, $R=e^{\int \left(\frac{1}{1+{\tan }^{2}z}\right){\sec }^{2}zdz}$

$$\begin{gathered} \Rightarrow R=e^{\int \left(\frac{1}{{\sec }^{2}z}\right){\sec }^{2}zdz}\left[\because \mathbf{1}\mathbf{+}{\mathbf{tan}}^{\mathbf{2}}\mathit{z}\mathbf{=}{\mathbf{sec}}^{\mathbf{2}}\mathit{z}\right] \\ \Rightarrow R=e^{\int dz} \\ \Rightarrow R=e^z \\ \Rightarrow R=e^{{\tan }^{-1}x}\left[\because x=\tan z\right] \end{gathered}$$

Therefore, the integrating factor of the differential equation $\left(1+x^2\right)\frac{dy}{dx}+y=e^{{\tan }^{-1}x}$ is $e^{{\tan }^{-1}x}$, so, option (C) is the correct answer.