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Need solution for RD Sharma Maths Class 12 Chapter 21 Differential Equation Excercise 21.10 Question 27

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Answer:  2 y \cos x=\cos 2 x+C

Hint: To solve this equation we use  \frac{d y}{d x}+P y=Q  where P,Q  are constants.

Give:  \begin{aligned} & \frac{d y}{d x}=y \tan x-2 \sin x \\ & \end{aligned}

Solution:  d x=\frac{d y}{d x}-y \tan x=-2 \sin x

\begin{aligned} &=\frac{d x}{d y}+P y=Q \\ &P=-\tan x \text { and } Q=-2 \sin x \end{aligned}


If  of differential equation is

 \begin{aligned} &I f=e^{\int P d x} \\ & \end{aligned}

=e^{-\int \tan x d x} \\

=e^{\log \cos x}

\begin{aligned} &=\cos x \quad\left[e^{\log x}=x\right] \\ & \end{aligned}

y I f=\int \text { QIf } d x+C \\

=y \cos x=-\int 2 \sin x \cos x d x+C \\

=y \cos x=-\int \sin 2 x d x+C \\

=y \cos x=\frac{\cos 2 x}{2}+C \\

=2 y \cos x=\cos 2 x+C

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