How to find the general solution to this ODE... Q:
Find the general solution to the ODE:$$t^2y''-ty'+y=0~~~~~~~~(*)$$given that $y_1=t~~$is a solution.
My intuitive solution:
Let $x=ln(t)$, then $y'=\frac{dy}{dx}\frac{1}{t}$ and $y''=\frac{d^2y}{dx^2}\frac{1}{t^2}-\frac{dy}{dx}\frac{1}{t^2}$
Plug into $(*)$ we get: $$t^2(\frac{d^2y}{dx^2}\frac{1}{t^2}-\frac{dy}{dx}\frac{1}{t^2})-t(\frac{dy}{dx}\frac{1}{t})+y=0$$$$\rightarrow~~\frac{d^2y}{dx^2}-2\frac{dy}{dx}+y=0$$ So the characteristic equation for this is $r^2-2r+1=0~\rightarrow~(r-1)^2=0~\rightarrow~r=1$.
Then $y_1(x)=y_2(x)=e^x~\rightarrow~y_1(t)=y_2(t)=e^{ln(t)}=t.~$But now the Wronskian is certainly zero. So I tried to multiply a $t$ such that $y_2(t)=t^2.~$But $y'_2=2t~~$and$~~ y''_2=2.~$Substituting these into $(*)$ I found that this doesn't solve the equation. 
Could someone please give me a hint on how to find the general solution?

HINT
Consider the method of the reduction of order.
You know one of the solutions to the differential equation, which is: $y_1 = t$
Now, call the second solution $y_2$ where $y_2 = vy_1 = vt$, where $v$ is a function of $t$.
Taking the derivative, we get:
$$y_2 = vt$$
$$y_2' = v't + v$$
$$y_2'' = v''t + v' + v' = v''t + 2v'$$
Substituting this into our equation:
$$t^2(v''t + 2v') -t(v't + v) + vt = 0$$
$$t^3v'' + 2t^2v' - t^2v' - vt + vt =0,\ t^3v'' + 2t^2v' - t^2v' = 0$$
Note: Upon using this method the $y$ term has dropped out. This method ensures that, and now we can find our second solution:
$$t^3v'' + t^2v' = 0$$
Now, we let $w = v'$, $w' = v''$
It follows that:
$$t^3w' + t^2w = 0$$
$$w' = -\frac{w}{t}$$
$$\frac{dw}{w} = -\frac{dt}{t}$$
$$\ln w = -\ln t$$
$$w = -t,\ v' = -t,\ v = -\frac{t^2}{2}$$
Now, our second solution is $$y_2 = vt = -\frac{t^3}{2}$$
Now, our general solution is formed:
$$y(t) = c_1t + c_2\frac{t^3}{2}$$
