Q:

Tricky Differential Equations Questions

I am self studying differential equations, and ran into some problems I have never seen before.
The majority of problems I have solved have come int the form of $Ay''-by'+C = 0$. I clearly understand how to solve these, and arrive at the general solutions.
However, I am now encountering a problem like 
$2t^2y''-5ty'+6y = 0$ where they give a solution $y(t) = t^\frac32)$. 
Does this now mean I have to work backwards from before since we already have a solution given to us?
Likewise, how does the general form look different in the case when
$2t^2y'' - 5ty' + 6t = t^2 - 1$. I know this revolves around methods of undetermined coefficients, and have guessed $Ax^2+Bx+C$ and have differentiating it twice over for y'' and once over for y', but I still haven't organized my work together to arrive at a general solution.
Thank you

A:

Hint: The differential equation
$$2t^2y''-5ty'+6y = 0$$
or
$$4t^2y''-10ty'+12y = 0$$
known as Euler differential equation, which can be solved with substitution $x=\ln2t$. In this case when the right side is zero, the differential equation 
$$(at)^2y''+paty'+qy = 0$$
is homogeneous and with $x=\ln at$ is converted to
$$y''+\left(\dfrac{p}{a}-1\right)y'+\dfrac{p}{a^2}y = 0$$
then with characteristic equation
$$\lambda^2+\left(\dfrac{p}{a}-1\right)\lambda+\dfrac{p}{a^2}=0$$
we may find it's general solution. 

Edit:
After substitution $x=\ln 2t$, your equation is
$$y''-\dfrac72y'+3y=2e^{2x}-2\tag{*}$$
from characteristic equation, the general solution is
$$y_g=C_1e^{2x}+C_2e^{\frac32x}$$
and while the right side of equation (*) has item $e^{2x}$ which is one of the general solutions, then let $y=Axe^{2x}+B$ to find particular solution with undetermined coefficient method. After $A$ and $B$ were found, put $x=\ln 2t$ and return main independent variable. One may find further examples here.

