Question

Let $y=y\left(x\right)$ be the solution of the differential equation, ${(x^2+1)}^2\frac{dy}{dx}+2x(x^2+1)y=1suchthaty\left(0\right)=0$. If $\sqrt{a}y\left(1\right)=\frac{\mathrm{\pi }}{32}$, then the value of $a$ is:

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{16}$
• D. $1$

Answer 1

The correct option is C

$\frac{1}{16}$

Explanation for the correct answer:

Step-1 : Solution if linear differential equation:

Given, ${(x^2+1)}^2\frac{dy}{dx}+2x(x^2+1)y=1$

We can write:

$$\begin{gathered} {(x^2+1)}^2\frac{dy}{dx}+2x(x^2+1)y=1 \\ \Rightarrow \frac{dy}{dx}+\left(\frac{2x}{x^2+1}\right)y=\frac{1}{{(x^2+1)}^2} \\ I.F.=e^{\int \frac{2x}{1+x^2}dx} \\ \therefore I.F.=(1+x^2) \end{gathered}$$

On solving further:

$$\begin{gathered} y(1+x^2)=\int \frac{1}{{\left(1+x^2\right)}^2}.(1+x^2)dx \\ y(1+x^2)={\tan }^{-1}x+c......\left(1\right) \end{gathered}$$

Given, $y\left(0\right)=0$

From equation $\left(1\right)$: $c=0$

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

Step-2 :Calculating the value of $a$:

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

Putting $x=1$ in the above equation:

$$\begin{gathered} y\times \left(2\right)={\tan }^{-1}\left(1\right) \\ \Rightarrow y\times \left(2\right)=\frac{\mathrm{\pi }}{4} \\ y=\frac{\mathrm{\pi }}{8} \\ \frac{1}{4}.y=\frac{\mathrm{\pi }}{32} \\ \therefore \sqrt{a}=\frac{1}{4}\mathbf{[}\mathbf{Comapring}\mathbf{}\mathbf{with}\mathbf{}\sqrt{\mathbf{a}}\mathbf{.}\mathbf{y}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{\pi }}{\mathbf{32}}\mathbf{]} \\ \Rightarrow a=\frac{1}{16} \end{gathered}$$

Hence, the correct answer is option (C).