- #1

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}##