2 Cos X D Y D X + 4 Y Sin X = Sin 2 X , Given that Y = 0 When X = π 3 . - Mathematics

Sum

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

Solution

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

$y = \cos x - 2 \cos^2 x$

Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 59 | Page 146