Question

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

• A. $y{e}^{{\tan }^{-1}x}=e^{{\tan }^{-1}x}+c$
• B. $y{e}^{{\tan }^{-1}x}=\frac{{\left(e^{{\tan }^{-1}x}\right)}^2}{2}+c$
• C. $e^{{\tan }^{-1}x}={\left(e^{{\tan }^{-1}x}\right)}^2+c$
• D. $y{e}^{{\tan }^{-1}x}=\frac{{\left(e^{{\tan }^{-1}x}\right)}^2}{3}+c$

Answer 1

Answer: (B)

$y{e}^{{\tan }^{-1}x}=\frac{{\left(e^{{\tan }^{-1}x}\right)}^2}{2}+c$

$\left(1+x^2\right)\frac{dy}{dx}+y=e^{{\tan }^{-1}x}$

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

$P=\frac{1}{1+x^2}Q=\frac{e^{\tan -1x}}{1+x^2}$

$I.F=e^{\int pdx}$

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

$=e^{\tan -1x}$

$y\times e^{\tan -1x}=\int \frac{e^{\tan -1x}}{1+x^2}\times =e^{{\tan }^{-1x}}dx+c$

$\Rightarrow y{e}^{\tan -1x}=\int \frac{e^{2\tan -1x}}{1+x^2}\times =dx+c$

$Putting{\tan }^{-1}x=t$

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

$\Rightarrow y{e}^{{\tan }^{-1}x}=\int e^{2t}dt+c$

$\Rightarrow y{e}^{{\tan }^{-1}x}=\frac{1}{2}{e}^{2t}+c$

$\Rightarrow y{e}^{{\tan }^{-1}x}=\frac{1}{2}{e}^{2{\tan }^{-1}x}+c$

$\Rightarrow y{e}^{{\tan }^{-1}x}=\frac{{\left(e^{{\tan }^{-1}x}\right)}^2}{2}+c$