Solve the following differential equation: (1+x^2) dy dx +y= ^ -1… - Vidyadip

Question

Solve the following differential equation:
$(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}+\text{y}=\tan^{-1}\text{x}$

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Answer

Here, $(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}+\text{y}=\tan^{-1}\text{x}$
$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{1+\text{x}^2}=\frac{\tan^{-1}\text{x}}{1+\text{x}^2}$
It is a linear differential equation. Comparing the equation by,
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
$\text{P}=\frac{1}{1+\text{x}^2},\text{Q}=\frac{\tan^{-1}\text{x}}{1+\text{x}^2}$
I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\frac{1}{1+\text{x}^2}\text{dx}}$
$=\text{e}^{\tan^{-1}\text{x}}$
Solution of the equation is given by,
$\text{y}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{dx + C}$
$\text{y}\big(\text{e}^{\tan^{-1}\text{x}}\big)=\int\frac{\tan^{-1}\text{x}}{1+\text{x}^2}\text{e}^{\tan^{-1}\text{x}}\text{dx + C}$
Let $\tan^{-1}\text{x}=\text{t}$
$\frac{1}{1+\text{t}^2}\text{dx}=\text{dt}$
So,
$\text{ye}^{\text{t}}=\int\text{t}\times\text{e}^{\text{t}}\text{dt + C}$
$=\text{t}\times\int\text{e}^{\text{t}}\text{dt}-\int\big(1\times\text{e}^{\text{t}}\text{dt}\big)\text{dt + C}$
Using integration by parts
$\text{ye}^{\text{t}}=\text{te}^{\text{t}}-\text{e}^{\text{t}}+\text{C}$
$\text{y}=(\text{t}-1)\text{Ce}^{-\text{t}}$
$\text{y}=(\tan^{-1}\text{x}-1)+\text{Ce}^{-\tan^{-1}\text{x}}$

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