Question

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

Answer 1

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

$\frac{dy}{dx}+\frac{y}{1+x^2}=\frac{e^{ta{n}^{-1}x}}{1+x^2}$

It is linear differential equation of first order.

Comparing with standard linear differential equation.

$\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$

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

Integrating factor(IF) = $e^{\int Pdx}=e^{\int \frac{1}{1+x^2}dx}=e^{ta{n}^{-1}x}$

Solution of LDE,

$y.IF=\int IFQ\left(x\right)dx+C$

$y.e^{ta{n}^{-1}x}=\int e^{ta{n}^{-1}x}.\frac{e^{ta{n}^{-1}x}}{1+x^2}dx+C$

$y.e^{ta{n}^{-1}x}=\int \frac{(e^{ta{n}^{-1}x}{)}^2}{1+x^2}dx+C$---(1)

To solving $\int \frac{(e^{ta{n}^{-1}x}{)}^2}{1+x^2}dx$

Put $e^{ta{n}^{-1}x}=t$

or $e^{ta{n}^{-1}x}.\frac{1}{1+x^2}dx=dt$

Therefore, $\int \frac{e^{ta{n}^{-1}x}.e^{ta{n}^{-1}x}}{1+x^2}dx=\int tdt$

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

Substituting in (1),

$y.e^{ta{n}^{-1}x}=\frac{(e^{ta{n}^{-1}x}{)}^2}{2}+C.$