Solving an Euler differential equation

[member 587159]

Homework Statement

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

Homework Equations

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:

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

 

[Ray Vickson]

Homework Statement

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

Homework Equations

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:

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

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

 

[vela]

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

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

This doesn't look right.

 

[member 587159]

This doesn't look right.

Thanks for pointing out!

 

[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.? ​

 

[member 587159]

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.? ​

I just checked and I made no typo, unfortunately.

 

[epenguin]

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