Question

The integrating factor of the differential equation $(1+x^2){\displaystyle \frac{dy}{dx}}+y=e^{{\tan }^{-1}x}$is?
A. ${\tan }^{-1}x$
B. $(1+x^2)$
C. $e^{{\tan }^{-1}x}$
D. ${\log }_e(1+x^2)$

Answer 1

Hint:Since the problem is based on a differential equation hence, it is necessary to check whether the equation is given, is in general form or not, and if it is not, then convert it into general form. And, then we will use the required formula $e^{\int p(x)dx}$ for finding out the integrating factor of a given problem and to get the accurate solution.

Formula Used:
$\int {\displaystyle \frac{1}{1+x^2}}dx={\tan }^{-1}x$
$IF=e^{\int p(x)dx}$

Complete step by step Solution:
The differential equation given is $(1+x^2){\displaystyle \frac{dy}{dx}}+y=e^{{\tan }^{-1}x}$ … (1)
Divide eq. (1) by $(1+x^2)$ in both LHS and RHS, we get
${\displaystyle \frac{dy}{dx}}+{\displaystyle \frac{y}{(1+x^2)}}={\displaystyle \frac{e^{{\tan }^{-1}x}}{(1+x^2)}}$
Comparing the above expression with the general form ${\displaystyle \frac{dy}{dx}}+p(x).y=g(x)$, we get
$$p(x)={\displaystyle \frac{1}{(1+x^2)}}$$ and $$g(x)={\displaystyle \frac{e^{{\tan }^{-1}x}}{(1+x^2)}}$$
Now, we know that Integrating Factor (IF) is calculated as: -
$IF=e^{\int p(x)dx}$
Also, we know that $\int {\displaystyle \frac{1}{1+x^2}}dx={\tan }^{-1}x$
$$IF=e^{\int {\displaystyle \frac{1}{1+x^2}}dx}=e^{{\tan }^{-1}x}$$
Thus, the Integrating Factor of given differential equation is $$e^{{\tan }^{-1}x}$$.

Hence, the correct option is C.

Note: In the problems based on differential equations, apply the required formula to get the accurate solution to the problem. Also, it is advised to use this method only in the equations which can be reduced to the general form i.e., ${\displaystyle \frac{dy}{dx}}+p(x).y=g(x)$. The Calculation part must be done very carefully.