Differential equation $x^2y''-xy'+2y=x\ln{x}$

Question

Hello there I am trying to solve $$x^2y''-xy'+2y=xe^x$$ I thought to use Laplace transform after rewriting as $$(-x)^2y''+(-x)y'+2y=-(-x)e^x$$thus $$F''(s)+F'(s)+2F(s)=-\left(\frac{1}{s-1}\right)'$$ but now I dont know how to solve: $$F''(s)+F'(s)+2F(s)=\frac{1}{(s-1)^2}$$ Taking $z''+z'+2z=0\,$ For the homogeneous equation I have found $$r^2+r+2=0$$ giving $$r_{1,2}=\frac{-1 \pm i\sqrt{7}}{2}$$ thus $$z_0=e^{-\frac{1}{2}p}\left(c_1\cos{\frac{\sqrt{7}}{2}p}+c_2\sin{\frac{\sqrt{7}}{2}p}\right)$$ I would appreciate some help with this. Edit: It was supposed to be: $$x^2y''-xy'+2y=x\ln{x}$$ But I have no ideea how to do the Laplace of the RHS now.

Answer 1

I'm solving the equation in your edit and not in the initial text.

Laplace transform is not needed here (you don't have initial conditions). The equation is much easier than it looks. If you didn't know, this is a Cauchy-Euler equation, which can be solved using the substitution $x = e^t$:

$$ \frac{d^2y}{dt^2} - 2\frac{dy}{dt} + 2y = te^t $$

The homogeneous solution is solved by the characteristic equation

$$ r^2 - 2r + 2 = 0 \implies r = 1 \pm i $$ $$ \implies y_h = c_1e^t\cos t + c_2e^t\sin t $$

The particular solution can be solved using undetermined coefficients

$$ y_p = (At+B)e^t \implies y_p = te^t $$

Putting it together

\begin{align} y &= te^t + c_1e^t\cos t + c_2e^t\sin t \\ &= x\ln x + c_1x\cos(\ln x) + c_2x\sin(\ln x) \end{align}

Answer 2

I assume that if you want to use Laplace Transform is because you have the initial conditions $\ y(0)=0\ $ and $\ y'(0)=0\ $

If I was you, I'd avoid LT cause is not so easy like in constants terms ODEs.

The easiest way is visualizing that the ODE has a No Homogeneous Cauchy Euler Form

Let $\ y=x^m\ $ then do the derivatives and substitute in the Homogeneous Part of the ODE

$$y_h(x)=x[C_1\cos(\ln|x|))+C_2\sin(\ln|x|)]$$

Apply Variation of Parameters where $y_1=x\cos(\ln|x|)\ $,$\ y_2=x\sin(\ln|x|)$ and the $f(x)$ for the Wronskians $\ W=x\ $, $\ W_1=-\ln|x|\sin(\ln|x|)\ $, $\ W_2=\ln|x|\cos(\ln|x|)\ $ is

$$f(x)=\frac{\ln|x|}{x}$$

Then $$u_1=\int\frac{W_1}{W}\qquad u_2=\int\frac{W_2}{W}$$ $$y_p(x)=y_1u_1+y_2u_2\\ y_p(x)=-x\ln|x|$$

$$y(x)=x\left\{C_1\cos(\ln|x|))+C_2\sin(\ln|x|)-\ln|x|\right\}$$

I hope anyone can solve it by LT like you want it.

If is a IVP centered in 0, is undefined for the $\ln|x|$ and you can't evaluate it unless you make an numerical aproximation, thus LT doesn't work.

Answer 3

Note that with the substitution $x=e^t$, $u(t)=y(e^t)$ on the equatio in the title, the equation becomes a linear DE with constant coefficients $$ \ddot u-2\dot u+2u=te^t. $$ That should be easy to solve with what ever method, Laplace transform, undetermined coefficients etc.