Let y=y(x) be the solution of the differential equation, (x^{2}+1)^{2}\frac{dy}{dx}+2x(x^{2}+1)y=1 such that y(0)=0. If \sqrt{a}\; y\; (1)=\frac{\pi}{32}, then the value of 'a' is :
 

  • Option 1)

    \frac{1}{16}

  • Option 2)

    \frac{1}{4}

  • Option 3)

    1

     

  • Option 4)

    \frac{1}{2}

 

Answers (1)

\\(x^{2}+1)^{2}\frac{dy}{dx}+2x(x^{2}+1)y=1\\\; \; \; \\y(0)=0

\frac{dy}{dx}+\frac{2xy}{x^{2}+1}=\frac{1}{(x^{2}+1)^{2}}

this is a linear differential equation with I.F=e^{\int \frac{2x}{x^{2}+1}dx}=e^{\ln(x^{2}+1)}=x^{2}+1

\therefore Solution \; y.(x^{2}+1)=\int (x^{2}+1)\frac{1}{(x^{2}+1)^{2}}dx=\int \frac{dx}{x^{3}+1}

=\tan^{-1}\; x+c

y(0)=0\Rightarrow 0=\tan^{-1}(0)+1=0\Rightarrow c=0

y(x^{2}+1)=\tan^{-1}(x)

y(x)=\frac{\tan^{-1}(x)}{x^{2}+1}

\sqrt{a} \; y(1)=\frac{\pi}{32}

We have y(1)=\frac{\tan^{-1}(1)}{2}=\frac{\frac{\pi}{4}}{2}=\frac{\pi}{8}

\frac{\sqrt{a}\; y(1)}{y(1)}=\frac{\frac{\pi}{32}}{\frac{\pi}{8}}\Rightarrow \sqrt{a}=\frac{8}{32}=\frac{1}{4}

a=\frac{1}{16}

 


Option 1)

\frac{1}{16}

Option 2)

\frac{1}{4}

Option 3)

1

 

Option 4)

\frac{1}{2}

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