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Need solution for RD Sharma Maths Class 12 Chapter 21 Differential Equation Excercise 21.10 Question 28

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Answer: $y e^{\sin x}=(\sin x-1) e^{\sin x}+C$

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

Give: $\begin{aligned} &\frac{d y}{d x}+y \cos x=\sin x \cos x \\ & \end{aligned}$

Solution: $\frac{d y}{d x}+y \cos x=\sin x \cos x$

$\begin{aligned} &=\frac{d x}{d y}+P y=Q \\ &P=\cos x \text { and } Q=\sin x \cos x \end{aligned}$

$If$ of differential equation is

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

$=e^{\int \cos x d x} \\$

$=e^{\sin x} \\$

$=\sin x \quad\left[e^{\sin x}=\sin x\right] \\$

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

$\begin{aligned} &y e^{\sin x}=\sin x \cos x e^{\sin x} d x \quad[\sin x=t, \cos x d x=d t] \\ & \end{aligned}$

$=\int t e^{t} d t \\$

$=t e^{t}-\int e^{t}+C \\$

$=t e^{t}-e^{t}+C \\$

$=e^{t}(t-1)+C \\$

$=y e^{\sin x}=(\sin x-1) e^{\sin x}+C$

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