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Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21.10 question 30

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Answer:  y \sin x=\frac{2}{3} \sin ^{3} 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} &(\sin x) \frac{d y}{d x}+y \cos x=2 \sin ^{2} x \cos x \\ & \end{aligned}

Solution:  \sin x \frac{d y}{d x}+y \cos x=2 \sin ^{2} x \cos x

\begin{aligned} &=\frac{d y}{d x}+\frac{y \cos x}{\sin x}=\frac{2 \sin ^{2} x \cos x}{\sin x} \\ & \end{aligned}

=\frac{d y}{d x}+y \cot x=2 \sin x \cos x \ldots(i) \\

=\frac{d y}{d x}+P y=Q \\

P=\cot x, Q=2 \sin x \cos x

If  of differential equation is

\begin{aligned} &\text { If }=e^{\int P d x} \\ & \end{aligned}

=e^{\int \cot x d x} \\

=e^{\log |\sin x|} \\

=\sin x \\

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

\begin{aligned} &=y \sin x=\int 2 \sin ^{2} x \cos x d x \\ & \end{aligned}

=2 \int \sin ^{2} x \cos x d x                           \quad[\sin x=t, \cos x d x=d t] \\

=2 \int t^{2} d t \\


\begin{aligned} &=\frac{2}{3} t^{3}+C \\ & \end{aligned}

=\frac{2}{3} \sin ^{3} x+C \\

y \sin x=\frac{2}{3} \sin ^{3} x+C

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