Euler ODE - particular solution problem

Question

I have this ODE: $$x^2y'' -xy' +y = 6x\ln(x)$$

Once I solve the homogenous part I get that $$y_h = xC_1 + x\ln(x)C_2$$

But I am having problems with the particular solution. When I try to find it using variation of constants:

$$C_1'x + C_2'x\ln(x)=0$$ $$C_1' + C_2'(1+\ln(x))=6x\ln(x)$$

The Wronskian is $x$ and if I solve the system of equations I get that $C_1'=-6x\ln^2(x)$ and $C_2' = 6x\ln(x)$

If I integrate both and plug them in the upper solution and add them, I get $$\frac{3x^3(\ln(x)+1)}{2} + C_1x + C_2x\ln(x)$$

The solution provided by Wolfram Alpha is $$C_1x + C_2x\ln(x)+x\ln^3(x)$$

Where did I go wrong? I solved the system of equations and I checked it on symbolab, I got $C_1$ and $C_2$ correctly. What did I do wrong? Thanks!

Answer 1

You got the homogeneous and Wronskian correct.

We have

$$y'' - \dfrac{y'}{x} + \dfrac{y}{x^2} = \dfrac{6 \ln(x)}{x}$$

Using Variation of Parameters (notice that the form to apply VoP is $y'' + \ldots$), from the homogeneous solution and ODE, we have

$$y_1 = x, y_2 = x \ln(x), f(x) = \dfrac{6 \ln(x)}{x}$$

So, we get

$$\begin{align} w_1 &= -\int \dfrac{f(x) y_2}{W}~dx = -\int \dfrac{6 \ln^2(x)}{x}~dx = - 2 \ln^3(x) \\ w_2 &= \int \dfrac{f(x) y_1}{W}~dx = \int \dfrac{6 \ln(x)}{x}~dx = 3 \ln^2(x) \end{align}$$

The solution is given by

$$y(x) = y_h(x) + y_1 w_1(x) + y_2 w_2(x)$$