Finding 2nd solution of second order ODE Function $x_1(t)=e^t$ is a solution of: $$tx''-(2t+1)x'+(t+1)x=0$$
Find second lineary independent solution.
I tried to do it the usual way, substituting $x=e^{\alpha t}$. But that gives me the only one solution that has been given already. I know the second solution will probably be $te^t$, but I dont know how to prove.it
 A: Function $x_1(t)=e^t$ is a solution of: $$tx''-(2t+1)x'+(t+1)x=0$$
Let us try the reduction of order approach, which is to try a second solution $x_2(t)=g(t)x_1(t)$. Inserting into the original DE, we have
$$
t(g(t)x_1(t))''-(2t+1)(g(t)x_1(t))'+(t+1)(g(t)x_1(t))=0
$$
The derivatives are 
$$
(g(t)x_1(t))' = g'(t)x_1(t) + x_1'(t)g(t)
$$
$$
(g(t)x_1(t))'' = g''(t)x_1(t) + 2g'(t)x_1'(t) + x_1''(t)g(t)
$$
Substituting gives
$$
g(t)\left(tx_1''(t)-(2t+1)x_1'(t)+(t+1)x_1(t)\right) + F(x_1,g,t)
=0
$$
where
$$
F(x_1,g,t) = t(g''(t)x_1(t) + 2g'(t)x_1'(t))-(2t+1)g'(t)x_1(t)
$$
The term in parentheses two lines above is zero because $x_1$ satisfies the original DE, leaving
$$
F(x_1,g,t)=0
$$
$$
t(g''(t)x_1(t) + 2g'(t)x_1'(t))-(2t+1)g'(t)x_1(t) = 0
$$
Now since, $x_1=x_1'=e^t$, that term cancels everywhere, leaving
$$
t(g''(t) + 2g'(t))-(2t+1)g'(t) = 0
$$
There are two $2tg'$ terms that also cancel, leaving
$$
tg''(t)-g'(t) = 0
$$
Letting $h=g'$ reduces this to a first order DE:
$$
th'(t)-h(t) = 0
$$
Separation gives
$$
\frac{1}{h(t)}\frac{dh}{dt} = \frac{1}{t}
$$
Now integrating $dt$
$$
\int \frac{1}{h} dh = \int \frac{1}{t} dt,
$$
$$
\log(h) = \log(t)+c,
$$
for some constant $c$. Exponentiating both sides gives
$$
h = Ct,
$$
For some other constant $C=e^c$. Since $g'=h$, we integrate $dt$ once more to get $g$
$$
g = At^2+B,
$$
where $A=\frac{C}{2}$ and $B$ is another constant of integration. Now we put together our final solution
$$
x_2(t) = (At^2+B)e^t
$$
Plugging this into the DE verifies that it is a second solution. You can verify it's linear independence using the Wronkskian determinant:
$$
\left|
\begin{array}{cc}
 e^t & e^t \left(A t^2+B\right) \\
 e^t & e^t \left(A t^2+B\right)+2 At e^t \\
\end{array}
\right|
 = 
2 A t e^{2 t},
$$
which does not identically vanish on any interval; so the solutions are independent.
A: If you have found the solutions $Cf(t)$, a good attempt to find more solutions is to vary the constant by looking at functions of the form $g(t)f(t)$ and solving for $g$.
When simplifying the equation for $g$, many terms will disappear since $f$ is a solution.
In your case, try to find a solution of the form $g(t)e^t$ and see what you get for $g$.
A: $$tx''-(2t+1)x'+(t+1)x=t(x'-x)'-(t+1)(x'-x)=0$$
$$ty'-(t+1)y=0$$
$$y'-(1+\frac1t)y=0$$
$$e^{\int-1-\frac1tdt}=e^{-t-lnt}=\frac1{te^t}$$
$$\frac{y'}{te^t}-\frac{t+1}{t^2e^t}y=(\frac y{te^t})'=0$$
$$y=ate^t$$
$$x'-x=ate^t$$
$$e^{-t}x'-e^{-t}x=(e^{-t}x)'=at$$
$$\frac x{e^t}=a_2t^2+b$$
$$x=a_2t^2e^t+be^t$$
