finding the solution geral Consider the equation
$$ty''-(t-1)y'+y=t^2e^{2t} $$ (eq.1)
Knowing that $$ y_1(t) = e^t $$ 
is a general solution of the corresponding homogeneous differential equation, determine the general solution of the homogeneous equation.
Checking the solution:
$$ te^t-(t-1)e^t+e^t=0$$
$$ 0=0 $$
Making
$$ y=uy_1 = ue^t$$
$$y'=e^tu+e^tu' $$
$$ y''= e^tu+u'e^t+u''e^t+u'e^t$$
Substituting in equation 1
$$u''te^t+u'(te^t+e^t)+2ue^t=0$$
Putting $$v=u'$$
$$v'(te^t)+v(te^t+e^t)+2u=0$$
simplifying
$$v't+v(t+1)+2u=0$$
Now I don't know how to continue.
Can you help?
Thanks
 A: $$ty''-(t-1)y'+y=t^2e^{2t}$$
$e^t$ is not a solution of the homogeneous solution that's why trying to reduce the order of the equation didn't work.Maybe you made a sign error and the equation is :
$$ty''-(t+1)y'+y=t^2e^{2t}$$
Here you can try reduction of order method. With $y=ve^t$.
$$\implies tv''+v'(t-1)=t^2e^t$$
Susbtitute $u=v'$ and solve as a first order DE. Or write it as:
$$\left (\frac {v'}t\right )'+\frac {v'}t=e^t$$
$$\left (e^t\frac {v'}t\right )'=e^{2t}$$
Integrate and substitute back $y=ve^t$
A: Assuming the DE as
$$
t y''-(t+1)y'+y=t^2e^{2t}
$$
here $y_1 = e^t$ is a solution for the homogeneous DE. It is easy to find also a polynomial solution for the homogeneous DE with $y_2 = t+1$ so 
$$
y_h = c_1 (t+1)+ c_2 e^t
$$
now making $y_p = c_1(t)(t+1)+c_2(t)e^t$ after substitution in the complete DE we arrive at
$$
-(t^2+1)c_1'+t(t+1)c_1'' +e^t(t-1)c_2'+t e^2c_2''-t^2e^{2t}=0
$$
here $c_1(t),c_2(t)$ are independent functions so we choose
$$
\cases{-(t^2+1)c_1'+t(t+1)c_1''= 0\\ e^t(t-1)c_2'+t e^tc_2''- t^2e^{2t}=0}
$$
or
$$
\cases{-(t^2+1)c_1'+t(t+1)c_1''= 0\\ (t-1)c_2'+t c_2''=t^2e^t}
$$
and now making $u=c_1', v= c_2'$ we have
$$
\cases{-(t^2+1)u+t(t+1)u'= 0\\ (t-1)v+t v'=t^2e^t}
$$
and then choosing $u=0$, $v = \frac 12 t e^t$ we have, (we are choosing a particular solution so we select the convenient value for the integration constants) 
$$
\cases{c_1(t) = 0\\ c_2(t) = \frac 12(t-1)e^t
}
$$
and finally
$$
y = y_h+y_p =  c_1 (t+1)+ c_2 e^t + 0(t+1)+\frac 12(t-1)e^t e^t
$$
or
$$
y =  c_1 (t+1)+ c_2 e^t +\frac 12(t-1)e^{2t}
$$
