Question

2 cos $x\frac{dy}{dx}+4y$ sin x = sin 2x, given that y = 0 when x = $\frac{\mathrm{\pi }}{3}$.

Answer 1

We have,
$2\cos x\frac{dy}{dx}+4y\sin x=\sin 2x$
$$\begin{gathered} \Rightarrow \frac{dy}{dx}+4y\frac{\sin x}{2\cos x}=\frac{2\sin x\cos x}{2\cos x} \\ \Rightarrow \frac{dy}{dx}+2y\tan x=\sin x \end{gathered}$$
$$\begin{gathered} \mathrm{Comparing}\mathrm{with}\frac{dy}{dx}+Py=Q,\mathrm{we}\mathrm{get} \\ P=2\tan x \\ Q=\sin x \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{2\int \tan xdx} \\ =e^{2\log \left(se\mathrm{c}x\right)} \\ ={\sec }^{2}x \\ \mathrm{So},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{given}\mathrm{by} \\ y\times \mathrm{I}.\mathrm{F}.=\int Q\times \mathrm{I}.\mathrm{F}.dx+C \\ \Rightarrow y{\sec }^{2}x=\int \sin x{\sec }^{2}xdx+C \\ \Rightarrow y{\sec }^{2}x=\int \tan x\sec xdx+C \\ \Rightarrow y{\sec }^{2}x=\sec x+C \\ \Rightarrow y=\cos x+C{\cos }^{2}x.....\left(1\right) \\ \mathrm{Now}, \\ \mathrm{When}x=\frac{\mathrm{\pi }}{3},y=0 \\ \therefore 0=\cos \frac{\mathrm{\pi }}{3}+C{\cos }^2\frac{\mathrm{\pi }}{3} \\ \Rightarrow 0=\frac{1}{2}+C\frac{1}{4} \\ \Rightarrow C=-2 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ y=\cos x-2{\cos }^{2}x \end{gathered}$$