General solution to a non-homogeneous Cauchy-Euler differential equation

Question

Here is the problem: Solve $ t^2x'' + x = \sin(\ln t)$ for $t>0$

I first find the fundamental set of solutions to the corresponding homogeneous equation $$ t^2x'' + x = 0$$
Let $$\ x(t)=t^r , x''(t) = r(r-1)t^{r-2} $$
Then $$ \ r(r-1)t^r + t^r = 0$$
$$ r^2 - r +1 = 0 \Rightarrow r = \frac{1\pm\sqrt{3}i}{2} $$
$$ \Rightarrow t^{\frac{1}{2}(1+\sqrt{3}i)} = e^{\frac{1}{2}\ln{t}}[\cos(\frac{\sqrt{3}}{2}\ln{t})+i \sin(\frac{\sqrt{3}}{2}\ln{t})]$$
$$ \Rightarrow x_1(t) = e^{\frac{1}{2}\ln{t}}\cos(\frac{\sqrt{3}}{2}\ln{t}) ;\ x_2(t)=e^{\frac{1}{2}\ln{t}}\sin(\frac{\sqrt{3}}{2}\ln{t})$$

(1) Is it necessary to check that the Wronskian for $x_1(t)\ \text{and}\ x_2(t)$ is non-zero?
(2) How do I proceed with finding the particular solution for the non-homogeneous case? What should my ansatz be?

Thank you!

Answer 1

Question 2:

$$t^2x'' + x = \sin(\ln t)$$ Substitute $u= \ln t$ $$t^2\left(\dfrac {1}{t^2}x''_u-\dfrac 1{t^2}x'_u \right)+x=\sin u$$ $$x''-x'+x =\sin u$$ Try for the particular solution: $$x_p=A\cos u$$

Answer 2

(1) It is blatantly obvious that $x_1$ and $x_2$ are linearly independent; I don't know why anybody would insist on checking a Wronskian in a case such as this.

Hint for (2): try for a particular solution of the form $x = a \sin(\ln(t)) + b \cos(\ln(t))$.

BTW: you might note that $e^{\frac{1}{2} \ln t} = t^{1/2}$.