# Solving an Euler differential equation

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1. May 26, 2017

### Math_QED

1. The problem statement, all variables and given/known data

Solve the differential equation $(2x+1)^2y'' + (4x+2)y' - 4y = x^2$
Can someone verify whether my solution is correct?

2. Relevant equations

3. The attempt at a solution

We perform the substitution $t = \ln|2x+1|$. Then, $e^t = |2x+1|$ and $x = \pm(e^t -1)/2$

Without loss of generality, let's solve the equation for $x = (e^t-1)/2$. (the other case is analogue)

We find $\frac{dy}{dx} = 2e^{-t}\frac{dy}{dt}$

$\frac{d^2y}{dx^2} = -4e^{-2t}\frac{dy}{dt}+4e^{-2t}\frac{d^2y}{dt^2}$

Hence, the differential equation becomes:

$16y'' - 16y = (e^t-1)^2$

This is a linear differential equation with constant coefficients:

The homogeneous is given by:

$y_h = c_1e^t + c_2e^{-t}$

Let's apply variation of the constants to find $y_p$, the particular solution.

$y_p = c_1(t)e^t + c_2(t)e^{-t}$

Then, we can find the functions $c_1'$ and $c_2'$ as solutions of the system:

$\begin{cases}c_1'(t)e^t + c_2'(t)e^{-t} = 0\\c_1'(t)e^t - c_2'(t)e^{-t} = (e^t-1)^2\end{cases}$

$\iff c_1'(t) = -c_2'(t) = \frac{(e^t-1)^2}{2e^t}$

Hence:

$c_1(t) = e^t/2 - t - e^{-t}/2 = -c_2(t)$ (if we set the constant of integration to $0$)

Hence, the solution is given by:

$y = y_p + y_h = c_1e^t + c_2e^{-t} -1 + 1/2e^{2t}-te^t + 1/2e^{-2t}+te^{-t}$

And by substituting back:

$y(x) = c_1(2x+1) + c_2\frac{1}{2x+1} + \frac{(2x+1)^2}{2} - \ln(2x+1)(2x+1) - 1 + \frac{1}{2(2x+1)^2} + \frac{\ln(2x+1)}{2x+1}$

2. May 26, 2017

### Ray Vickson

What is preventing you from putting your formula for $y(x)$ into the DE to test if it works?

3. May 26, 2017

### vela

Staff Emeritus
This doesn't look right.

4. May 26, 2017

### Math_QED

Thanks for pointing out!

5. May 27, 2017

### epenguin

What a pity that last LHS term is –4y instead of 4y, in which case the LHS would have been an exact second derivative. Sure you copied it out right.?

6. May 27, 2017

### Math_QED

I just checked and I made no typo, unfortunately.

7. May 27, 2017

### epenguin

Pity they didn't ask good question. I can't see any way to hammer that one into something so nice.