Question

Solution of the differential equation \(\frac{dy}{dx} -y \,tan⁡x = \frac{sin⁡x cos^2⁡x}{y^2}\)  is ______

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

(b) \(y^2 cos^2 x = \frac{sin^6 x}{2} + c \)

(c) \(y^3 cos^3 x = \frac{-cos^6 x}{2} + c  \)

(d) \(y^4 cos^5 x = \frac{sinx cos⁡x}{2} + c \)

This question was addressed to me in quiz.

I'm obligated to ask this question of Bernoulli Equations in portion Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Correct option is (c) \(y^3 cos^3 x = \frac{-cos^6 x}{2} + c  \)

To explain I would say: \(\frac{dy}{dx} -y \,tan⁡x = \frac{sin⁡x cos^2⁡x}{y^2}\)  multiplying by

\(y^2 \rightarrow y^2  \frac{dy}{dx} – y^3  tan⁡x =  sin⁡x \,cos^2⁡x \)

put \(t = y^3\rightarrow3y^2 \frac{dy}{dx} = \frac{dt}{dx} \,or\, y^2 \frac{dy}{dx} = \frac{1}{3} \frac{dt}{dx}\)

substituting \(\frac{dt}{dx}-3t \,tan⁡x = 3sin⁡x \,cos^2⁡x \)

this equation is linear in t i.e it is of the form \(\frac{dt}{dx} + Pt = Q, e^{\int P \,dx} = e^{\int -3tan⁡x  \,dx}\)

\(e^{-3 log⁡\, sec⁡x} = cos^3 \,x\) its solution is \(te^{\int P \,dx} = \int Q \,e^{\int P \,dx} dx + c\)

t cos^3 x=∫3sin⁡x cos^2x    cos^3 x dx + c = ∫3sin⁡x cos^5⁡x dx+c, put v=cos x

dv=-sin x dx i.e \(\int 3sin⁡x \,cos^5\, ⁡x \,dx = \int -3v^5 \,dx = \frac{-v^6}{2} = \frac{-cos^6 x}{2}\) hence its solution becomes \(t cos^3 x = \frac{-cos^6 x}{2} + c \rightarrow y^3 cos^3 \,x = \frac{-cos^6 x}{2} + c.\)