Question

Find the general solution of the D.E 2y-4xy’-log y’=0.

(a) \(y(p) = \frac{2c}{p} – 1 + \frac{log⁡p}{2} \)

(b) \(y(p) = \frac{c}{2p} – 2 + log⁡p\)

(c) \(x(p) = \frac{-1}{p} + \frac{c}{p^2} \)

(d) \(x(p) = \frac{1}{2p} + \frac{c}{p^{1/2}} \)

I had been asked this question during an interview for a job.

The origin of the question is Clairaut’s and Lagrange Equations in chapter Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

The correct choice is (a) \(y(p) = \frac{2c}{p} – 1 + \frac{log⁡p}{2} \)

To explain: Let y’=p and hence given equation is in Lagrange equation form

i.e 2y=4xp+log p …..(1) differentiating both sides of the equation

2dy=4xdp+4pdx+\(\frac{dp}{p}\) and dy=pdx

–> 2pdx=4xdp+4pdx+\(\frac{dp}{p}\) –> -2pdx=4pdx+\(\frac{dp}{p}\)

-2p\(\frac{dx}{dp} = 4x + \frac{1}{p} \rightarrow \frac{dx}{dp} + \frac{2}{p} x = \frac{-1}{2p^2}\) (p≠0)…….this is a linear D.E for the function x(p)

I.F is \(e^{\int \frac{2}{p} \,dp} = e^{log⁡p^2}  = p^2\) and solution is x(p) \(p^2 = \int p^2 *\frac{-1}{2p^2} \,dp + c\)

\(x(p) = \frac{-1}{2p} + \frac{c}{p^2}\)  substituting back in (1) we get \(2y=4p\left(\frac{-1}{2p} + \frac{c}{p^2}\right) + log p\)

\(y(p) = \frac{2c}{p} – 1 + \frac{log⁡p}{2} \).