Question

The solution of $\frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac{1}{{(1+x^2)}^2}$ given that $x=1,y=0$ is

• A. $y(1+x^2)={\tan }^{-1}x-\frac{\pi }{4}$
• B. $y(1-x^2)={\tan }^{-1}x-\pi /4$
• C. $y(1+x^2)={\tan }^{-1}x+\frac{\pi }{4}$
• D. $y(1-x^2)={\tan }^{-1}x+\pi /4$

Answer 1

The correct option is A $y(1+x^2)={\tan }^{-1}x-\frac{\pi }{4}$

$\frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac{1}{{\left(1+x^2\right)}^2}itisfirstlinearD.E,herep=\frac{2x}{1+x^2},q=\frac{1}{{\left(1+x^2\right)}^2}I.F=e^{\int pdx}=e^{\int \frac{2x}{1+x^2}dx}=e^{\log \left(x^2+1\right)}=x^2+1SolutionofD.Eisy\times I.F=\int Q\times I.Fdx+C\Rightarrow y\times \left(x^2+1\right)=\int \frac{1}{{\left(x^2+1\right)}^2}\left(x^2+1\right)dx+C\Rightarrow \left(x^2+1\right)y=\int \frac{1}{x^2+1}dx+C\Rightarrow \left(x^2+1\right)y={\tan }^{-1}x+CGiven:y=0whenx=1So,0={\tan }^{-1}x+C\Rightarrow 0=\frac{\pi }{4}+c\Rightarrow C=-\frac{\pi }{4}thereforesolutionofD.Eis\left(x^2+1\right)y={\tan }^{-1}x-\frac{\pi }{4}$