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$\frac{d y}{d x}+2 y \tan x=\sin x ; y=0$ when $x=\frac{\pi}{3}$


The given differential equation is $\frac{d y}{d x}+2 y \tan x=\sin x$.

This is a linear equation of the form:

$\frac{d y}{d x}+p y=Q$ (where $p=2 \tan x$ and $Q=\sin x$ )

$\frac{d y}{d x}+p y=Q($ where $p=2 \tan x$ and $Q=\sin x)$

Now, I.F $=e^{\int p d x}=e^{\int 2 \tan x d x}=e^{2 \log |\sec x|}=e^{\log \left(\sec ^{2} x\right)}=\sec ^{2} x$

The general solution of the given differential equation is given by the relation,

$y(\mathrm{I.F} .)=\int(\mathrm{Q} \times \mathrm{I.F}) d x+\mathrm{C}$

$\Rightarrow y\left(\sec ^{2} x\right)=\int\left(\sin x \cdot \sec ^{2} x\right) d x+\mathrm{C}$

$\Rightarrow y \sec ^{2} x=\int(\sec x \cdot \tan x) d x+\mathrm{C}$

$\Rightarrow y \sec ^{2} x=\sec x+\mathrm{C}$          …(1)

Now, $y=0$ at $x=\frac{\pi}{3}$.


$0 \times \sec ^{2} \frac{\pi}{3}=\sec \frac{\pi}{3}+\mathrm{C}$

$\Rightarrow 0=2+\mathrm{C}$

$\Rightarrow \mathrm{C}=-2$

Substituting C = –2 in equation (1), we get:

$y \sec ^{2} x=\sec x-2$

$\Rightarrow y=\cos x-2 \cos ^{2} x$

Hence, the required solution of the given differential equation is $y=\cos x-2 \cos ^{2} x$.


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