Solving an inhomogeneous Euler-Cauchy differential equation (3rd order) I have the following differential equation that I need to solve:
$x^3 y^{(3)} + 3 x^2 y'' + x y' = x^3 ln(x)$
I have managed to find the homogeneous solution which is:
$y_h = c_1 + c_2 ln\ |x| + \frac{1}{2}c_3ln^2\ |x| $
But I don't know how to calculate the particular solution.
I tried to substitute $x = e^t$ but that gives me some garbage answer.
 A: try this here as a particular solution $$y_p=Ax^3+Bx^3\log(x)$$
A: $$x^3y'''(x)+3x^2y''(x)+xy'(x)=x^3\ln(x)\Longleftrightarrow$$

Let $v(x)=y'(x)$, which gives $y''(x)=v'(x)$ and $y'''(x)=v''(x)$:

$$x^3v''(x)+3x^2v'(x)+xv(x)=x^3\ln(x)\Longleftrightarrow$$
$$x^2v''(x)+3xv'(x)+v(x)=x^2\ln(x)\Longleftrightarrow$$

The general solution will be the sum of the complementary solution
and particular solution.
Find the complementary solution by solving:

$$x^2v''(x)+3xv'(x)+v(x)=0\Longleftrightarrow$$

Assume a solution, proportional to $x^\mu$ for some constant $\mu$.
Substitute $v(x)=x^\mu$:

$$x^2\cdot\frac{\text{d}^2}{\text{d}x^2}\left(x^\mu\right)+3x\cdot\frac{\text{d}}{\text{d}x}\left(x^\mu\right)+x^\mu=0\Longleftrightarrow$$

Substitute $\frac{\text{d}^2}{\text{d}x^2}\left(x^\mu\right)=(\mu-1)\mu x^{\mu-2}$ and $\frac{\text{d}}{\text{d}x}\left(x^\mu\right)=\mu x^{\mu-1}$:

$$x^2(\mu-1)\mu x^{\mu-2}+3x\mu x^{\mu-1}+x^\mu=0\Longleftrightarrow$$
$x^{\mu}(\mu^2+2\mu+1)=0\Longleftrightarrow$$

Assuming $x\ne0$, zeros must come from the polynomial:

$$\mu^2+2\mu+1=0\Longleftrightarrow$$
$$(\mu+1)^2=0\Longleftrightarrow$$
$$\mu=-1$$
Now, find the particular solution:
$$x^2v''(x)+3xv'(x)+v(x)=x^2\ln(x)$$
List the basis solutions in $v_c(x)$ so $v_{c_1}(x)=\frac{1}{x}$ and $v_{c_2}(x)=\frac{\ln(x)}{x}$.
Compute the Wronskian of $v_{c_1}(x)$ and $v_{c_2}(x)$:
$$\mathcal{W}(x)=\left|\begin{matrix}
  \frac{1}{x} & \frac{\ln(x)}{x} \\
  \frac{\text{d}}{\text{d}x}\left(\frac{1}{x}\right) & \frac{\text{d}}{\text{d}x}\left(\frac{\ln(x)}{x}\right)
 \end{matrix}\right|=\left|\begin{matrix}
  \frac{1}{x} & \frac{\ln(x)}{x} \\
  -\frac{1}{x^2} & \frac{1}{x^2}-\frac{\ln(x)}{x^2}
 \end{matrix}\right|=\frac{1}{x^3}$$
Divide the differential equation by the leading term's coefficient $x^2$:
$$v''(x)+\frac{3v'(x)}{x}=\frac{v(x)}{x^2}=\ln(x)$$
Let:


*

*$$q(x)=\ln(x)$$

*$$r_1(x)=-\int\frac{q(x)v_{c_2}(x)}{\mathcal{W}(x)}\space\text{d}x=-\int x^2\ln^2(x)\space\text{d}x=\frac{x^3\left(9\ln^2(x)-6\ln(x)+2\right)}{27}+\text{K}_1$$

*$$r_2(x)=\int\frac{q(x)v_{c_1}(x)}{\mathcal{W}(x)}\space\text{d}x=\int x^2\ln(x)\space\text{d}x=\frac{x^3\left(3\ln(x)-1\right)}{9}+\text{K}_2$$


So:
$$v(x)=v_c(x)+r_1(x)v_{c_1}(x)+r_2(x)v_{c_2}(x)$$
Now substitute back $v(x)=y'(x)$:
$$y(x)=\int\left(v_c(x)+r_1(x)v_{c_1}(x)+r_2(x)v_{c_2}(x)\right)\space\text{d}x$$
