Question

`2 cos x(dy)/(dx)+4y sin x = sin 2x," given that "y = 0" when "x = pi/3.`

Answer 1

We have,

`2 cos x(dy)/(dx)+4y sin x = sin 2x`

\[\Rightarrow \frac{dy}{dx} + 4y\frac{\sin x}{2 \cos x} = \frac{2\sin x \cos x}{2 \cos x}\]

\[ \Rightarrow \frac{dy}{dx} + 2y \tan x = \sin x\]

\[\text{Comparing with} \frac{dy}{dx} + Py = Q,\text{ we get}\]

\[P = 2\tan x\]

\[Q = \sin x\]

Now,

\[I . F . = e^{2\int\tan x dx} \]

\[ = e^{2\log\left( sec x \right)} \]

\[ = \sec^2 x\]

So, the solution is given by

\[y \times I . F . = \int Q \times I . F . dx + C\]

\[ \Rightarrow y \sec^2 x = \int\sin x \sec^2 x dx + C\]

\[ \Rightarrow y \sec^2 x = \int\tan x \sec x dx + C\]

\[ \Rightarrow y \sec^2 x = \sec x + C\]

\[ \Rightarrow y = \cos x + C \cos^2 x . . . . . \left( 1 \right)\]

Now,

\[\text{When }x = \frac{\pi}{3}, y = 0 \]

\[ \therefore 0 = \cos \frac{\pi}{3} + C \cos^2 \frac{\pi}{3}\]

\[ \Rightarrow 0 = \frac{1}{2} + C\frac{1}{4}\]

\[ \Rightarrow C = - 2\]

Putting the value of C in (1), we get

\[y = \cos x - 2 \cos^2 x\]