Question

Sovle the differential equation: $(1+x^2)\frac{dy}{dx}+y=e^{ta{n}^{-1}x}$

Answer 1

Given $(1+x^2)\frac{dy}{dx}+y=e^{ta{n}^{-1}x}\Rightarrow \frac{dy}{dx}+\frac{1}{1+x^2}y=\frac{e^{ta{m}^{-1}}x}{1+x^2}$

This is linear differential equation of the form $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$

So, $P\left(x\right)=\frac{1}{1+x^2},Q\left(x\right)=\frac{e^{ta{n}^{-1}x}}{1+x^2}$.

Now, I.F =$e^{{\int }_{1+x^2}^{1}dx}=e^{ta{n}^{-1}x}$

$\therefore$ required solution is : $y\left(e^{ta{n}^{-1}x}\right)=\int e^{ta{n}^{-1}x}\frac{e^{ta{n}^{-1}x}}{1+x^2}dx+C$

$\Rightarrow y\left(e^{ta{n}^{-1}x}\right)=\int tdt+C[\text{Put}{e}^{ta{n}^{-1}x}=t\Rightarrow \frac{e^{ta{n}^{-1}x}}{1+x^2}dx=dt]$

$\Rightarrow y\left(e^{ta{n}^{-1}x}\right)=\frac{t^2}{2}+C\Rightarrow y\left(e^{ta{n}^{-1}x}\right)=\frac{1}{2}{e}^{2ta{n}^{-1}x}+C$

$\therefore y=\frac{1}{2}{e}^{ta{n}^{-1}x}+C{e}^{ta{n}^{-1}x}$ is the required solution.