Second order homogeneous differential equation with non-constant coefficients How can I solve a 2nd order differential equation with non-constant coefficients like the following?
$$ty''-(t+1)y'+y=0$$
If I'm not wrong, I have only seen methods (apart from the reduction of order method) for finding a solution when the coefficients are constant. How can I do this?
 A: If you use the Laplace transform technique, then you can reduce the order of the ode as

$$ (s-s^2)Y'(s) + (2-3s)Y(s) + c = 0, $$

where $Y(s)$ is the Laplace transform of $y(t)$. Now, solving the first order ode gives

$$ Y(s) =  \frac{c_1\,s^2+c_2}{s^2(s-1)}. $$

Taking the inverse Laplace transform gives the solution of the original ode

$$ y(t) = A(t+1)+B e^{t}, $$

where $A, B$ are arbitrary constants.
Notes:

*

*The Laplace transform is given by


$$ Y(s) = \int_{0}^{\infty} y(t) e^{-st} dt. $$



*To find the inverse Laplace transform, you can use partial fraction as


$$ Y(s) = -{\frac {c_{{2}}}{s}}+{\frac {c_{{1}}+c_{{2}}}{s-1}}-{\frac {c_{{2}}}{
{s}^{2}}}.$$

A: Hint: The equation $ty''-(t+1) y'+y=0$ can be rearranged as:
$$ty''-(t+1) y'+y=0$$
$$0=ty''-(t+1)y'+y=ty''-ty'-y'+y=t(y''-y')-(y'-y)\\
\Leftrightarrow t(y''-y')=(y'-y)\\
\Leftrightarrow t\dfrac{d}{dt}(y'-y)=(y'-y).$$
Substituting $u=y'-y$, this becomes the 1st order ODE $$tu'=u\\
\Leftrightarrow\dfrac{u'}{u}=\dfrac{1}{t}\\
\Leftrightarrow\int\dfrac{u'}{u}dt=\int\dfrac{du}{u}=\int\dfrac{1}{t}dt\\
\Leftrightarrow\log{|u|}=\log{|t|}+C\\
\Leftrightarrow u(t)=c_1t$$
Once you solve for $u(t)$, the problem reduces to solving the linear 1st order ODE with constant coefficients, $y'-y=u(t)=c_1 t$.
$$y'-y=u(t)=c_1 t\\
\implies e^{-t}y'-e^{-t}y=u(t)=c_1 t e^{-t}\\
\implies \dfrac{d}{dt}\left(e^{-t}y\right)=c_1 t e^{-t}\\
\implies e^{-t}y = c_2 + c_1\int t e^{-t} dt\\
\implies e^{-t}y = c_2 + c_1 e^{-t}(t+1)\\
\implies y(t) = c_2 e^{t} + c_1 (t+1)$$
