I have been given the question:
Using y1 and y2 from part (a), use variation of parameters to find the general solution to $$2t^2y''-5ty'+3y = \frac{5t^3}{1+t^{5/2}}, t>0$$ For full marks, you must check that you have found the solution.
Here $y_1=t^3$ and $y_2=t^{1/2}$. If V.o.P states that $y=u_1y_1+u_2y_2$, I came out with values of $$u_1=ln(t^{5/2})-ln(t^{5/2}+1) +C_1$$ and $$u_2=ln(t^{5/2}+1) +C_2$$ which gives a general solution of $$t^3[ln(t^{5/2})-ln(t^{5/2}+1) +C_1]+t^{1/2}[ln(t^{5/2}+1) +C_2]$$ Is this correct? And if so, how would I go about 'checking I have found the solution'?