Question

Find the particular solution of the differential equation \(\frac{dy}{dx}=\frac{9y \,log⁡x}{5x \,log⁡y}\).

(a) (log⁡y)^2+(log⁡x)^2=0

(b) (log⁡y)^2-(log⁡x)^2=0

(c) log⁡y-log⁡x=0

(d) 2 log⁡x+log⁡y=0

This question was addressed to me in a job interview.

Enquiry is from Methods of Solving First Order & First Degree Differential Equations in division Differential Equations of Mathematics – Class 12

Answer 1

Right choice is (b) (log⁡y)^2-(log⁡x)^2=0

Easy explanation: \(\frac{dy}{dx}=\frac{9y \,log⁡x}{5x \,log⁡y}\)

Separating the variables, we get

\(\frac{5 \,log⁡y}{y} dy=\frac{9 \,log⁡x}{x} dx\)

Integrating both sides, we get

\(5\int \frac{log⁡y}{y} \,dy=9\int \frac{log⁡x}{x} \,dx\) –(1)

First, for integrating \(\frac{log⁡y}{y}\)

Let log⁡y=t

Differentiating w.r.t y, we get

\(\frac{1}{y}\) dy=dt

∴\(\int \frac{log⁡y}{y} \,dy=\int t \,dt\)

=\(\frac{t^2}{2}=\frac{log⁡y^2}{2}\)

Similarly integrating \(\frac{log⁡x}{x}\)

Let log⁡x=t

Differentiating w.r.t x, we get

\(\frac{1}{x}\) dx=dt

∴\(\int \frac{log⁡x}{x} dy=\int t \,dt\)

=\(\frac{t^2}{2}=\frac{(log⁡x)^2}{2}\)

Hence, equation (1), becomes

\(\frac{(log⁡y)^2}{2}=\frac{(log⁡x)^2}{2}+C\) –(2)

Given y=2, we get x=2

Substituting the values in equation (2), we get

\(\frac{(log⁡y)^2}{2}=\frac{(log⁡x)^2}{2}+C\)

C=0

Therefore, the equation becomes \((log⁡y)^2=(log⁡x^2)\)

∴(log⁡y)^2-(log⁡x)^2=0.