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Provide Solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 18

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Answer:   y \sin x=x^{2} \sin x+C

Hint: To solve this we convert \cot x  to \frac{\cos x}{\sin x}  formula.

Give:  \frac{d y}{d x}+\cot x y=x^{2} \cot x+2 x

Solution:  \begin{aligned} & \frac{d y}{d x}+P y=Q\\ & \end{aligned}

P=\cot x, Q=x^{2} \cot x+2 x\\

I f=e^{\int P d x}\\

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

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

=\sin x

\begin{aligned} &=y I f=\int Q I f d x+C \\ & \end{aligned}

=y \sin x=\int\left(x^{2} \cot x+2 x\right) \sin x d x+C \\

=y \sin x=\int x^{2} \frac{\cos x}{\sin x} \sin x+2 x \sin x d x+C \\

=y \sin x=\int x^{2} \cos x d x+\int 2 x \sin x d x+C

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

{\left[\int u v d x=u \int v d x-\int \frac{d y}{d x} \int v d x d x\right]} \\

=x^{2} \sin x-2 \int x \sin x d x+2 \int x \sin x d x+C \\

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

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