Question

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

Answer 1

$(1+x^2)\frac{dy}{dx}=e^{m{\tan }^{-1}x}-y⟶1$
$\frac{dy}{dx}=\frac{e^{m{\tan }^{-1}x}}{(1+x^2)}-\frac{y}{1+x^2}$
$\frac{dy}{dx}+\frac{y}{1+x^2}=\frac{e^{m{\tan }^{-1}x}}{(1+x^2)}⟶2$
Comparing it with general linear differential equation
$\frac{dy}{dx}+yP\left(x\right)=Q\left(x\right)$
$\therefore P\left(x\right)=\frac{1}{1+x^2}$ & $Q\left(x\right)=\frac{e^{m{\tan }^{-1}x}}{1+x^2}$
Integrating factor, IF$=e^{\int P\left(x\right)dx}$
$P\left(x\right)=\frac{1}{1+x^2}$
$\int P\left(x\right)dx=\int \frac{1}{1+x^2}dx={\tan }^{-1}x$
IF$=e^{{\tan }^{-1}x}⟶3$
Multiply the differential equation by $e^{{\tan }^{-1}x}$, we get
$e^{{\tan }^{-1}x}\frac{dy}{dx}+\frac{y{e}^{{\tan }^{-1}x}}{1+x^2}=\frac{e^{{\tan }^{-1}x.(m+1)}}{1+x^2}$
$e^{{\tan }^{-1}x}\frac{dy}{dx}+y\frac{d}{dx}\left(e^{{\tan }^{-1}x}\right)=\frac{e^{(m+1){\tan }^{-1}x}}{1+x^2}$
Using product rule we get,
$\frac{d}{dx}\left({ye}^{{\tan }^{-1}x}\right)=\frac{e^{(m+1){\tan }^{-1}x}}{1+x^2}$
Integrating, $y{e}^{{\tan }^{-1}x}=\int \frac{e^{(m+1){\tan }^{-1}x}}{1+x^2}dx$
Put $ta{n}^{-1}x=t\Rightarrow \frac{dx}{1+x^2}=dt$
$y{e}^{{\tan }^{-1}x}=\int e^{(m+1)t}dt=\frac{e^{(m+1)t}}{m+1}+C$