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प्रश्न
`2 cos x(dy)/(dx)+4y sin x = sin 2x," given that "y = 0" when "x = pi/3.`
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उत्तर
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
\[y = \cos x - 2 \cos^2 x\]
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