I keep trying to solve this but I end up needing to do integration by parts like 3 or 4 times. My only question is, is that going to be the only way to do this? If so it will literally take me hours. I don't need a solution, just a confirmation that if the requirement is to use variation of parameters, then will I need to take integrals that need to be done by IBP?
Thank you and sorry if this is not a good question.
Hint.
This is a linear DE so it's solution can be written as
$$ y = y_h + y_p\\ y''_h-25y_h=0\\ y''_p-25y_p=x $$
The solution for $y_h = c_1 e^{5x}+c_2 e^{-5x}$.
Assuming now
$$ y_p = c_1(x)e^{5x}+c_2(x)e^{-5x} $$
substituting into the complete DE we have
$$ e^{5 x} \left(C_1'(x)+10 C_1(x)\right)+e^{-5 x} \left(C_2'(x)-10 C_2(x)\right)-x = 0 $$
with $C_1(x) = c_1'(x), C_2(x) = c_2'(x)$
Now choosing and solving
$$ C_1'(x)+10 C_1(x)=0\\ e^{-5 x} \left(C_2'(x)-10 C_2(x)\right)-x = 0 $$
we have
$$ C_1(x) = K_1e^{-10x} = c_1'(x)\\ C_2'(x)-10 C_2(x)=x e^{5x} $$
This last equation in $C_2(x)$ can be solved applying again the variation of constants method etc.
NOTE
$$ c_1'(x) = K_1 e^{-10x}\\ c_2'(x) = K_2e^{10x}-\left(\frac x5+\frac{1}{25}\right)e^{5x} $$
etc.