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  • Need solution for RD Sharma Maths Class 12 Chapter 21 Differential Equation exercise 21.5 question 15.

Need solution for RD Sharma Maths Class 12 Chapter 21 Differential Equation exercise 21.5 question 15.

Answers (1)

Answer: y=\left (2x+\sin \frac{2x}{2} \right )-3x+2\sin x-\log \left |\sec x-\tan x \right |+C

Hint: You have to integrate by applying integration of \cos x.

Given: \cos x \frac{dy}{dx}-\cos 2x=\cos 3x

Solution: \cos x \frac{dy}{dx}-\cos 2x=\cos 3x

\begin{aligned} &d y=\frac{\cos 3 x+\cos 2 x}{\cos x} d x \\ &d y=\frac{\left(4 \cos ^{3} x-3 \cos x+2 \cos ^{2} x-1\right)}{\cos x} d x \end{aligned}

dy=(4cos^2x-3+2\cos x-\sec x) dx

Integrate on both sides

d y=\int 4 \cos ^{2} x d x-\int 3 d x+\int 2 \cos x d x-\int \sec x d x

y=\left (2x+\sin \frac{2x}{2} \right )-3x+2\sin x-\log \left |\sec x-\tan x \right |+C

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