Differential Equation Past Paper Question 
a) Find the general solution to the differential equation: $$2\frac{d^{2}y}{dx^{2}}-4\frac{dy}{dx}+20y=0.$$ b) Find the general solution to the differential equation: $$x^{2}\frac{d^{2}y}{dx^{2}}+4x\frac{dy}{dx}+2y=\frac{1}{x}.$$

I am having trouble with these two differential equations in a past paper I am going through. 
Thanks in advance for any replies! 

I'll just try to expand on the comments.
These two equations are linear ODEs, the first one being homogeneous.
To solve the first one, guess the solution ofthe form $y = e^{ax}$ and solve for $a$. You should get two values of $a$, say $a_1$ and $a_2$, hence two solutions: $e^{a_1x}$ and $e^{a_2x}$. The theory of ODEs will tell you that for $x \in \mathbb R$, there are at most $2$ linearly independent solutions, so you know that $c_1 e^{a_1 x} + c_2 e^{a_2 x}$ is your complete solution. Well, you must know that $e^{a_1 x}$ and $e^{a_2 x}$ are linearly independent before making the last conclusion. This is always the case when $a_1 \ne a_2$ (which is true here).
To solve the second one, first solve its homogeneous version: $x^2 \frac{d^2 y}{dx^2} + 4x \frac{dy}{dx} + 2y = 0$. This can be solved by first guessing the form of solution $y = x^a$ and solving for $a$. (This is equivalent to setting $x = e^t$ in Robert's comment.) You'll get two values for $a$, say $a_1$ and $a_2$, hence two solutions $x^{a_1}$ and $x^{a_2}$. The theory of ODEs, again, tells you that for $x > 0$, $y = c_1 x^{a_1} + c_2 x^{a_2}$ is the complete solution to the homogeneous equation. To deal with the non-homogeneous term, use variation of parameters: Assume $y = u_1(x)x^{a_1} + u_2(x)x^{a_2}$. Solve the system
$$
\begin{pmatrix}
x^{a_1} & x^{a_2} \\
a_1 x^{a_1 - 1} & a_2 x^{a_2 - 1}
\end{pmatrix}
\begin{pmatrix}
u_1'(x) \\ u_2'(x)
\end{pmatrix} =
\begin{pmatrix}
0 \\
1/x^3
\end{pmatrix}
$$
for $u_1'$ and $u_2'$, then integrate to get $u_1$ and $u_2$ (keep the two constants that appear). Substitute back into $y = u_1(x)x^{a_1} + u_2(x)x^{a_2}$ to get your final answer.
