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

Answers (1)

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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