# Euler ODE - particular solution problem

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!

• Did you integrate $-6 \ln^2(x)/x$ and $6 \ln(x)/x$ because you divide out the Wronskian, but also divide by the leading $x^2$? In fact, to get the ODE in the right form, it is better to write the ODE as $$y'' - \dfrac{y'}{x} + \dfrac{y}{x^2} = \dfrac{6 \ln(x)}{x}$$
– Moo
Commented Jul 17, 2020 at 22:36

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

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

• +1 nice moo ... Commented Jul 17, 2020 at 22:49
• I never knew I had to divide the equation by the leading coefficient. Why is that? Truth be told, I used this method for solving ODEs with constant coefficients (no x variable multiplying the derivatives of y). Commented Jul 17, 2020 at 23:34
• @johndoe: Look at the form in the blue box at tutorial.math.lamar.edu/classes/de/VariationofParameters.aspx. Notice how it "must" be in the form $y''...$ for the generalized approach to work. That is what caused your issue and is a typical problem.
– Moo
Commented Jul 17, 2020 at 23:44
• I absolutely never knew about this. I could have easily brushed it off as a typo in my workbook because it's prone to having mistakes. Thanks though! Commented Jul 18, 2020 at 0:00