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Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 27 maths textbook solution.

Answers (1)

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

Hint : To solve this equation we use \frac{dy}{dx}+Py=Q where P,Q are constants.

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

Solution : \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} \\ &=e^{-\int \tan x d x} \\ &=e^{\log \cos x} \end{aligned}

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

        \begin{aligned} &=y \cos x=\frac{\cos 2 x}{2}+C \\ &=2 y \cos x=\cos 2 x+C \end{aligned}

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