Solve $0=x-\dot x t - t^2 \ddot x, x(1)=0, x(2)=3$ Solve $0=x-\dot x t - t^2 \ddot x, x(1)=0, x(2)=3$
I have no idea how to do this DE.
Wolframalpha gives me $\frac{A(t^2+1)}{t}$ but I expect a 2 parameter family of solutions.
 A: Rearranging conveniently the given equation, one reaches the following result
\begin{align*}
& t^{2}x^{\prime\prime} + tx^{\prime} - x = 0 \Longleftrightarrow (t^{2}x^{\prime\prime} + 2tx^{\prime}) - (tx^{\prime} + x) = 0 \Longleftrightarrow (t^{2}x^{\prime})^{\prime} - (tx)^{\prime} = 0 \Longleftrightarrow\\\\
& t^{2}x^{\prime} - tx = k \Longleftrightarrow x^{\prime} - \frac{x}{t} = \frac{k}{t^{2}} \Longleftrightarrow \frac{x^{\prime}}{t} - \frac{x}{t^{2}} = \frac{k}{t^{3}} \Longleftrightarrow \left(\frac{x}{t}\right)^{\prime} = \frac{k}{t^{3}} \Longleftrightarrow\\\\
& \frac{x}{t} = -\frac{k}{2t^{2}} + c \Longleftrightarrow x(t) = -\frac{k}{2t} + ct
\end{align*}
Now it just remains to plug in the initial conditions.
A: This is an equation of the Euler-Cauchy type. Try finding basis solutions in the form $t^m$. This gives the characteristic equation $0=1-m-m(m-1)=1-m^2=(1-m)(1-m)$. So $t,t^{-1}$ are the basis solutions. $x(1)=0$ leads to $x(t)=A(t-t^{-1})$. The second condition then gives $A=2$.
