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(NONE) Given differential equation: $(1+x^2) \frac{dy}{dx} = e^{	an^{-1} x} - y$.
Divide by $(1+x^2)$: $\frac{dy}{dx} + \frac{y}{1+x^2} = \frac{e^{\

Vedclass · Accepted Answer

$y e^{	an^{-1} x} = \frac{e^{2 	an^{-1} x} + 1}{2}$

Vedclass · Answer

(NONE) Given differential equation: $(1+x^2) \frac{dy}{dx} = e^{	an^{-1} x} - y$.
Divide by $(1+x^2)$: $\frac{dy}{dx} + \frac{y}{1+x^2} = \frac{e^{	an^{-1} x}}{1+x^2}$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{1+x^2}$ and $Q = \frac{e^{	an^{-1} x}}{1+x^2}$.
Integrating Factor $(IF)$ = $e^{\int P dx} = e^{\int \frac{1}{1+x^2} dx} = e^{	an^{-1} x}$.
The solution is $y \cdot IF = \int Q \cdot IF dx + C$.
$y e^{	an^{-1} x} = \int \frac{e^{	an^{-1} x}}{1+x^2} \cdot e^{	an^{-1} x} dx + C$.
Let $t = 	an^{-1} x$,then $dt = \frac{1}{1+x^2} dx$.
$y e^{	an^{-1} x} = \int e^{2t} dt + C = \frac{e^{2t}}{2} + C = \frac{e^{2 	an^{-1} x}}{2} + C$.
Given $y=1$ when $x=0$: $1 \cdot e^0 = \frac{e^0}{2} + C \Rightarrow 1 = \frac{1}{2} + C \Rightarrow C = \frac{1}{2}$.
Thus,$y e^{	an^{-1} x} = \frac{e^{2 	an^{-1} x} + 1}{2}$.

Find the particular solution of the following differential equation,given that $y=1$ when $x=0$: $(1+x^2) \frac{dy}{dx} = e^{\tan^{-1} x} - y$.

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