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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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