Question

Let $y=y\left(x\right)$ be the solution of the differential equation $\sin x\left(\frac{dy}{dx}\right)+y\cos x=4x,x\in (0,\pi ).$ If $y\left(\frac{\mathrm{\pi }}{2}\right)=0$, then $y\left(\frac{\mathrm{\pi }}{6}\right)$ is equal to:

• A. $\left(-\frac{8}{9}\right){\mathrm{\pi }}^2$
• B. $\left(-\frac{4}{9}\right){\mathrm{\pi }}^2$
• C. $\left(\frac{4}{9\sqrt{3}}\right){\mathrm{\pi }}^2$
• D. $\left(-\frac{8}{9\sqrt{3}}\right){\mathrm{\pi }}^2$

Answer 1

The correct option is A

$\left(-\frac{8}{9}\right){\mathrm{\pi }}^2$

Explanation for the correct solution:

Step-1:Solution of linear differential eauation:

Given, $\sin x\left(\frac{dy}{dx}\right)+y\cos x=4x$

Solving the differential equation:

$$\begin{gathered} \sin x\left(\frac{dy}{dx}\right)+y\cos x=4x \\ \Rightarrow \frac{dy}{dx}+y\mathrm{co}tx=4x\cos ecx \\ I.F.=e^{\int cotxdx} \\ =\sin x \end{gathered}$$

On solving further, we get:

$$\begin{gathered} y\sin x=\int 4x.\cos ecx.\sin xdx \\ \Rightarrow y\sin x=\int 4xdx \\ \Rightarrow y\sin x=2x^2+C \end{gathered}$$

Step-2 :Calculating the value of $y\left(\frac{\mathrm{\pi }}{6}\right)$:

Since, $y\left(\frac{\mathrm{\pi }}{2}\right)=0$:

$$\begin{gathered} 0=2{\left(\frac{\mathrm{\pi }}{2}\right)}^2+C \\ C=-\frac{{\mathrm{\pi }}^2}{2} \\ \therefore y\sin x=2x^2-\frac{{\mathrm{\pi }}^2}{2} \end{gathered}$$

Therefore, for $y\left(\frac{\mathrm{\pi }}{6}\right)$, we have:

$$\begin{gathered} y\sin \left(\frac{\mathrm{\pi }}{6}\right)=2{\left(\frac{\mathrm{\pi }}{6}\right)}^2-\frac{{\mathrm{\pi }}^2}{2} \\ \Rightarrow \frac{y}{2}=\mathrm{\pi }\left(\frac{1}{18}-\frac{1}{2}\right) \\ \Rightarrow \mathrm{y}=-\frac{8{\mathrm{\pi }}^2}{9} \end{gathered}$$

Hence, the correct answer is option (A).