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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

Answers (1)

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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