Question

Solve the differential equation  \(\frac{dy}{dx} = \frac{x^2+y^2}{3xy}\) is _______

(a) xp=(x^2+2y^2)^-3

(b) x^2 p=(x^2-2y^2)^3

(c) x^4 p=(x^2-2y^2)^-3

(d) x^6 p=(x^2+2y^2)^3

This question was posed to me in a job interview.

Question is from Separable and Homogeneous Equations topic in division Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Correct choice is (b) x^2 p=(x^2-2y^2)^3

To elaborate: \(\frac{dy}{dx} = \frac{x^2+y^2}{3xy} = \frac{1+\frac{y^2}{x^2}}{3 \frac{y}{x}}\) we can clearly see that it is an homogeneous equation

hence substituting \(y = vx\rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx} = \frac{1+v^2}{3v}\)

separating the variables and integrating

\(x \frac{dv}{dx} = \frac{1+v^2}{3v} – v = \frac{1-2v^2}{3v}\)

\(\int \frac{3v}{1-2v^2} \,dv = \int \frac{1}{x} dx\)…….substituting 1-2v^2=t→-4v dv=dt we get

\(\frac{-3}{4}  log⁡ \,t = log⁡ \,x + log⁡ \,c   \rightarrow \frac{-3}{4} log⁡(1-2v^2) = log ⁡\,cx…..but \,v = \frac{y}{x}\)

\(-3log⁡(\frac{x^2-2y^2}{x^2})=4log⁡ \,cx \rightarrow log⁡(\frac{x^2-2y^2}{x^2})^{-3} = log⁡ \,kx^4 \)

\(\frac{x^6}{(x^2-2y^2)^3} = kx^4 \rightarrow x^2 \,p = (x^2-2y^2)^3\) is the solution where p is constant.