Solve differential equation: d y d x -y x=e^x - Vidyadip

Question

Solve differential equation: $\frac{d y}{d x}-y \tan x=e^x$

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Answer

The given differential equation is,
$\frac{d y}{d x}-y \tan x=e^x$
It is a linear differential equation.
Comparing it with $\frac{d y}{d x}+p y=Q$
$P=-\tan X, Q=e^{X}$
$\text { L.F. }=e^{\int pdx x}$
$=e^{-\int \tan x dx}$
$=e^{-\log \sec x}$
$=\cos x$
Solution of the given equation is given by,
$y \text { (I.F.) }=\int Q \times(I . F) d x+c_1$
$y \cos x=\int e^x \cos x d x+c_1$
$y \cos x=I+c_1 \ldots .(1)$
$I=\int e^x \cos x d x$
Using equation by parts
$I=e^x \int \cos x d x-\int\left(e^x \int \cos x d x\right) d x+c_2$
$=e^x \sin x-\int e^x \sin x d x+c_2$
$=e^x \sin x-\left[e^x\left(\int \sin x d x\right)-\int\left(e^x\left(\int \sin x d x\right) d x\right]+c_2\right.$
$I=e^x \sin x+e^x \cos x-\int e^x \cos x d x+c_2$
$I=e^x(\sin x+\cos x)-I+c_2$
$2 I=e^x(\sin x+\cos x)+c_2$
$I=\frac{1}{2} e^x(\sin x+\cos x)+c_3$
Using equation $(i)$
$y \cos x=\frac{1}{2} e^x(\sin x+\cos x)+c$

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