Question
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}=\text{y}\tan\text{x}-2\sin\text{x}$

Answer

Here,$\frac{\text{dy}}{\text{dx}}=\text{y}\tan\text{x}-2\sin\text{x}$
It is a linear differential equation. Comparing the equation by,
$\text{P}=-\tan\text{x},\text{Q}=-\sin\text{x}$
I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{-\int\tan\text{xdx}}$
$=\text{e}^{-\log\sec\text{x}}$
$=\frac{1}{\sec\text{x}}$
Solution of the equation is given by,
$\text{y}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{dx + C}$
$\frac{\text{y}}{\sec\text{x}}=\int-\frac{2\sin\text{x}}{\sec\text{x}}\text{dx+ C}$
$\text{y}\cos\text{x}=-\int2\sin\text{x}\cos\text{xdx + C}$
$\text{y}\cos\text{x}=-\int\sin2\text{xdx + C}$
$\text{y}\cos\text{x}=\frac{\cos2\text{x}}{2}+\text{C}$
$\text{y}=\frac{\cos2\text{x}}{2\cos\text{x}}+\frac{\text{C}}{\cos\text{x}}$

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