Second Order Nonhomogeneous Differential Equation.

Question

The Eulero equation of the functional

$$J[y]=\int_a^b \frac{y'^2}{x^3} \; \mathrm dx,$$

leads to the following differential equation:

$$\frac{y''}{x^3}-3\frac{y'}{x^4}=0.$$

I rewrote the equation as

$$xy''-3y'=0,$$

integrating by parts I got:

$$y'x-4y=0.$$

and by separation of variables:

$$y=c_1x+c_2$$

which is not a solution of the original differential equation. Where did I commit illegal steps?

Answer 1

You are making a mistake with logarithms. Separating variables you get $$ \begin{align} \frac{y'}{y} &= 4 \frac{1}{x} \\ \implies \mathrm{ln}(y) &= 4 \mathrm{ln}(x) +c \\ \implies \mathrm{ln}(y) &= \mathrm{ln}(x^4) +c \\ \implies y &= Dx^4 \end{align} $$

Answer 2

The general solution of $y'x-4y=0$ is given by

$y=cx^4$.