Question

Solution of the differential equation \(\frac{dy}{dx}\) + y cot ⁡x = cos⁡x is ______

(a) \(y cos \,x = \frac{sin^2 x}{2} + c\)

(b) \(y sin \,x = \frac{sin^2 x}{2} + c\)

(c) \(y sin \,x = \frac{cos^2 x}{4} + c\)

(d) \(y cos \,x = -\frac{sin^2 x}{4} + c\)

This question was addressed to me in my homework.

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

Answer 1

The correct choice is (b) \(y sin \,x = \frac{sin^2 x}{2} + c\)

To explain: \(\frac{dy}{dx}\) + y cot⁡x = cos⁡x 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=cot x, Q=cos x, \(e^{\int P \,dx} = e^{\int cot⁡x \,dx} = e^{log⁡sin⁡x} = sin x\)

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

y sin x=∫(cos x*sin x) dx + c….. substitute sin x=t to solve integral

\(y sin x = \frac{t^2}{2} + c = \frac{sin^2 x}{2} + c\) is the solution.