Question

Let $y=y\left(x\right)$ be the solution of the differential equation $\cos x\frac{dy}{dx}+2y\sin x=\sin 2x,x\in (0,\pi /2).$If $y\left(\frac{\mathrm{\pi }}{3}\right)=0$, then $y\left(\frac{\mathrm{\pi }}{4}\right)$ is equal to:

• A. $2+\sqrt{2}$
• B. $\sqrt{2}-2$
• C. $\frac{1}{\sqrt{2}}-1$
• D. $2-\sqrt{2}$

Answer 1

The correct option is B

$\sqrt{2}-2$

Explanation for the correct answer:

Calculating the value of $y\left(\frac{\mathrm{\pi }}{4}\right)$:

Given,

$\begin{array}{rcl}\cos x\frac{dy}{dx}+2y\sin x& =& \sin 2x\\ & \Rightarrow & \frac{dy}{dx}+\frac{2\sin x}{\cos x}y=2\sin x\left[\because \sin 2x=2\mathrm{sinxcosx}\right]\end{array}$

Now,

$\begin{array}{rcl}I.F.& =& e^{\int \frac{2\sin x}{\cos x}dx}\\ & =& e^{2\ln secx}\\ & =& se{c}^{2}x\end{array}$

Therefore,

$\begin{array}{rcl}y.se{c}^{2}x& =& \int 2\sin x.se{c}^{2}xdx\\ & \Rightarrow & y.se{c}^{2}x=\int 2\tan x.secxdx\\ & \Rightarrow & y.se{c}^{2}x=2secx+c\end{array}$

Substituting $x=\frac{\mathrm{\pi }}{3},y=0$ in the above equation:

$$\begin{gathered} \begin{array}{rcl}& \Rightarrow & 0=2sec\frac{\mathrm{\pi }}{3}+c\\ & \Rightarrow & c=-4\end{array} \\ \therefore yse{c}^{2}x=2secx-4 \end{gathered}$$

Now, putting $x=\frac{\mathrm{\pi }}{4}$:

$$\begin{gathered} \Rightarrow y.se{c}^2\left(\frac{\mathrm{\pi }}{4}\right)=2sec\left(\frac{\mathrm{\pi }}{4}\right)-4 \\ \Rightarrow y.2=2\sqrt{2}-4 \\ \Rightarrow y=\sqrt{2}-2 \end{gathered}$$

Hence the correct answer is option (B).