homogenous 2nd order ode with exponential in LHS I've come across a type of homogeneous 2nd order ode that I haven't seen before and isn't covered in the book I have. I've also tried googling it but haven't been able to find anything similar since I don't know the name for this type of problem. Any help in how to solve it would be much appreciated. Question is below.
Find general solutions with the replacement $x= e^t$ to the following differential equations, in each of which a function $y(x)$ is defined for $x>0$. Here, $y^{\prime}$ denotes the derivative of first order with respect to $x$ for a function $y(x)$.
1.)$$x^2y^{\prime\prime}-xy^{\prime}-x-3y=0$$
2.)$$x^3y^{\prime\prime\prime}+6x^2y^{\prime\prime}+4xy^{\prime}-4y=0$$
 A: Since you need to express as a function of $t$ using $x=e^t$ (that is to say $t=\log(x)$), start with $$\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=\frac 1 x\frac{dy}{dt}=e^{-t}\frac{dy}{dt}$$ Deriving a second time $$\frac{d^2y}{dx^2}=\frac d{dt}\big(\frac{dy}{dx}\big)\frac{dt}{dx}=\frac d{dt}\big(e^ {-t}\frac{dy}{dt}\big)\frac{dt}{dx}=e^{-t} \left(\frac{d^2y}{dt^2}-\frac{dy}{dt}\right)e^{-t}=e^{-2t} \left(\frac{d^2y}{dt^2}-\frac{dy}{dt}\right)$$ So, the differential equation $$x^2y^{\prime\prime}-xy^{\prime}-x-3y=0$$ write $$e^{2t}e^{-2t} \left(\frac{d^2y}{dt^2}-\frac{dy}{dt}\right)-e^te^{-t}\frac{dy}{dt}-e^t-3t=0$$ that is to say $$\frac{d^2y}{dt^2}-2\frac{dy}{dt}-3y=e^t$$ which looks simple.
Continue the same way to get $\frac{d^3y}{dx^3}$ for the second problem.
A: The key point point about these differential equations is that the coefficient of each derivative of order $n$ is $x^n$. I know that the second order ODE the you wrote is named Euler differential equation. You can use the special property I mentioned to find the general solution of the corresponding homogeneous ODE namely ${y_h}$ and then find a particular solution ${y_p}$ for the non-homogeneous ODE. Finally, summing the two solutions ${y_h} + {y_p}$ will give you the general solution of the ODE. I will work out the first example, leaving the second for you. we have
$${x^2}y'' - xy' - 3y = x$$
fist consider the corresponding homogeneous ODE
$${x^2}y'' - xy' - 3y = 0$$
Now we guess that $y(x) = {x^r}$ can be a solution. This is motivated by the special property of this ODE that was pointed out. Now put it into the homogeneous ODE to get
$$\left( {{r^2} - r - 3} \right){x^r} = 0$$
so $x^r$ can be a solution if and only if
$${r^2} - r - 3 = 0\,\,\,\,\,\, \to \,\,\,\,\,\,\,{r_{1,2}} = {{1 \pm \sqrt {13} } \over 2}$$
so we have
$${y_h} = {c_1}{x^{{r_1}}} + {c_2}{x^{{r_2}}}$$
for finding a particular solution since the RHS is a polynomial we consider
$${y_p} = Ax + B$$
putting into the non-homogeneous ODE leads to
$$0 - Ax - 3\left( {Ax + B} \right) =  - 4Ax - 3B = x\,\,\,\,\,\,\, \to \,\,\,\,\,\,\left\{ \matrix{
   - 4A = 1 \hfill \cr 
   - 3B = 0 \hfill \cr}  \right.\,\,\,\,\, \to \,\,\,\,\,\,\left\{ \matrix{
  A =  - {1 \over 4} \hfill \cr 
  B = 0 \hfill \cr}  \right.$$
and hence
$${y_p} =  - {1 \over 4}x$$
and the general solution of the whole problem will be
$$y = {y_h} + {y_p} = {c_1}{x^{{r_1}}} + {c_2}{x^{{r_2}}} - {1 \over 4}x$$
