Solve second order ODE knowing one of its solutions 
Solve $t^2x''-4tx'+6x=0$ knowing that $x_1(t)=t^2+t^3$ is a particular solution

So I assumed the general solution will be in form of $x(t)=C_1 x_1(t)+C_2 x_2(t)$ and $x_2 = v(t)x_1(t)$
So now I have to find $x_2$
$$x_2=v(t^2+t^3)$$
$$x_2'=v'(t^2+t^3)+v(2t+3t^2)$$
$$x_2''=v''(t^2+t^3)+2v'(2t+3t^2)+v(2+6t)$$
And if I try to plug it into my equation I get something very complicated, as almost nothing reduces. Is it the right way to do it?
EDIT: ill try to plug it in here and see what happens since I probably made a mistake
$$t^2(v''(t^2+t^3)+2v'(2t+3t^2)+v(2+6t))-4t(v'(t^2+t^3)+v(2t+3t^2))+6(v(t^2+t^3))=0$$
dividing both sides by $t^2$
$$v''(t^2+t^3)+2v'(2t+3t^2)+v(2+6t)-4(v'(t+t^2)+v(2+3t))+6(v(1+t))=0$$
$$v''(t^2+t^3)+v'(2(2t+3t^2)-4(t+t^2))+v((2+6t)-4(2+3t)+6(1+t))=0$$
$$v''(t^2+t^3)+v'(4t+6t^2-4t-4t^2)+v(2+6t-8-12t+6+6t)=0$$
$$v''(t^2+t^3)+v'(2t^2)=0$$
Yeah I made a mistake, thanks for pointing that out.
 A: You're close. If you think about it, your particular solution is actually two separate solutions. 
$$x_1(t) = t^2$$
and
$$x_2(t) = t^3$$
These two form a basis for all solutions. This type of ODE can be solved with the ansatz that $x(t) = t^p$ for some power $p$. (Because the derivatives are accompanied by powers of $t$.) So with this guess, you get 
$p(p-1) - 4p + 6 =0$
This means $p$ is two or three.
A: This is an "Euler equation". Doing the transformation 
$$
x(t)=z(\log t),
$$
you obtain for $z$ the equation
$$
z''-5z'+6z=0,
$$
and hence
$$
z(t)=c_1\mathrm{e}^{2t}+c_2\mathrm{e}^{3t},
$$
and hence
$$
x(t)=c_1 t^2+c_2t^3,
$$
which is the general solution of your equation.
A: In standard form, the equation is $x'' - \frac{4}{t}x' + \frac{6}{t^2}x = 0$. The Wronskian is then $W = c_1e^{-\int -\frac{4}{t}dt} = c_1t^4$. Then, since the Wronskian is $W = x_1x_2'-x_1'x_2$, we get a first order ODE.
$$
c_1t^4 = (t^2+t^3)x_2'-(2t+3t^2)x_2 \implies c_1t^3 = t(1+t)x_2'-(2+3t)x_2
$$
Again using the method of the integrating factor on that expression yields
$$
x_2 = c_2t^2(t+1)-c_1t^2
$$
Picking almost any $c_1$ or $c_2$ should give you a linearly independent solution. For example, with $c_2=0$ and $c_1=-1$, you get $x_2=t^2$.
Edit: here's a link that talks about the Wronskian in second order ODEs.
