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  • Provide Solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 19

Provide Solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 19

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Answer:  y \sec x=x^{2} \sin x+2 x \cos x-2 \sin x+C

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

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

Solution:  \frac{d y}{d x}+P y=Q \\

\begin{aligned} &P=\tan x, Q=x^{2} \cos ^{2} x \\ \end{aligned}

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

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

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

=\sec x

 \begin{aligned} &=y I f=\int Q I f d x+C \\ &=y \sec x=\int x^{2} \cos ^{2} x(\sec x) d x+C \\ &=y \sec x=\int x^{2} \cos ^{2} x(1 / \cos x) d x+C \\ &=y \sec x=\int x^{2} \cos x d x+C \\ &=y \sec x=x^{2} \int \cos x d x-\int(2 x \cos x d x) d x+C \end{aligned}

Using integration by parts

 \begin{aligned} &=y \sec x=x^{2} \sin x-2 \int x^{2} \sin x d x+C \\ &=y \sec x=x^{2} \sin x-2 \int x \sin x d x-\int(\sin x d x) d x+C \\ &=y \sec x=x^{2} \sin x-2 \int x \sin x d x-\int(\sin x d x) d x+C \\ &=y \sec x=x^{2} \sin x+2 x \cos x-2 \sin x+C \end{aligned}

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