Question

For the differential equation \(\frac{dy}{dx}\) – 3y cot⁡x = sin⁡2x; y=2 when x=\(\frac{\pi}{2}\), its particular solution is ______

(a) y = 2cos^2 x + 4cos^3 x

(b) y = -2sin^3 x + 4sin^2 x

(c) y = -2sin^2 x + 4sin^3 x

(d) y = 4cos^2 x + 2sin^3 x

I have been asked this question in examination.

This interesting question is from First Order Linear Differential Equations topic in division Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Correct option is (c) y = -2sin^2 x + 4sin^3 x

The explanation: \(\frac{dy}{dx}\) – 3y cot⁡x = sin⁡2x is of the form \(\frac{dy}{dx}\) + Py = Q where P & Q is a function of x

only given DE is linear DE in y here P=-3cot x, Q=sin 2x, \(e^{\int P \,dx} = e^{\int -3cot⁡x \,dx}\)

\( =e^{-3 log⁡sin⁡x} = e^{log⁡(\frac{1}{sin^3x})} = \frac{1}{sin^3x}\)

 Linear DE solution is given by \(ye^{\int P \,dx} = \int Q \,e^{\int P \,dx} dx + c\)

\(y \frac{1}{sin^3x} = \int sin \,2x \frac{1}{sin^3x}  \,dx + c = \int 2 \,sin⁡x \,cos⁡x \frac{1}{sin^3x} dx = \int 2 \,cos⁡x \frac{1}{sin^2x}  dx\)

substitute sin x=t to solve integral \(y \frac{1}{sin^3x} = \int 2 \frac{1}{t^2} \,dt + c = \frac{-2}{t} + c = \frac{-2}{sin⁡x} + c\)

it is given that when x=π/2, y=2→2=-2+c→c=4 its particular solution is thus given

 by \(y \frac{1}{sin^3x} = \frac{-2}{sin⁡x} + 4 \rightarrow y = -2sin^2 \,x + 4sin^3 \,x.\)