Question

The solution of the D.E $(1+x^2)\frac{dy}{dx}+y=e^{{\tan }^{-1}x}$ is :

• A. $2e^{{\tan }^{-1}x}=e^{{\tan }^{-1}x}+c$
• B. $2y{e}^{{\tan }^{-1}x}=e^{2{\tan }^{-1}x}+c$
• C. $2y{e}^{{\tan }^{-1}x}=e^{{\tan }^{-1}x}+c$
• D. $y{e}^{2{\tan }^{-1}x}=e^{{\tan }^{-1}x}+c$

Answer 1

The correct option is B $2y{e}^{{\tan }^{-1}x}=e^{2{\tan }^{-1}x}+c$

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

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

The above DE is of the form $\frac{dy}{dx}+P_y=Q$

Integrating factor $(I.F)=e^{\int pdx}$

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

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

Solution : $y(I.F.)=\int Q(I.F)dx$

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

Let $ta{n}^{-1}x=t$

$\frac{1}{1+x^2}dx=dt$

$=\int e^{2t}dt$

$=\frac{e^{2t}}{2}+c$

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

$\Rightarrow 2y{e}^{ta{n}^{-1}x}=e^{2ta{n}^{-1}x}+c$