Question

Solution of the differential equation \(\frac{dy}{dx} = \frac{3x-6y+7}{x-2y+4}\) is _____

(a) log(x-2y+4)^2=c

(b) 6x-2y+log(x-2y+2)^2=c

(c) x-2y+log(x-2y+6)^2=c

(d) -5x+log(x-2y+3)^2=c

The question was posed to me during an interview for a job.

This interesting question is from Reducible to Homogenous Form topic in division Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Correct choice is (b) 6x-2y+log(x-2y+2)^2=c

To explain: \(\frac{dy}{dx} = \frac{3x-6y+7}{x-2y+4} \rightarrow  \frac{dy}{dx} = \frac{3(x-2y)+7}{(x-2y)+4}\)…..here coefficient of x and y in the numerator & denominator are proportional hence substituting \(x-2y = t \rightarrow 1 – 2\frac{dy}{dx} = \frac{dt}{dx}\)

\(1- \frac{dt}{dx} = 2\left(\frac{3t+7}{t+4}\right) \rightarrow \frac{dt}{dx} = \frac{t+4-6t-14}{t+4} =\frac{-5(t+2)}{t+4}\)

separating the variables and hence integrating

\( \int \frac{t+4}{t+2} \,dt = \int -5 \,dx \rightarrow \int \left(1 + \frac{2}{t+2}\right) \,dt = -5x + c\)

t + 2 log(t+2) = -5x + c –> x – 2y + 2 log(x-2y+2) + 5x = c

6x-2y+log(x-2y+2)^2 = c is the solution.