if $f$ and $g$ are solutions of linear homogenous ODE, prove $af+bg$ is also a solution 
a) If $f$ and $g$ are solutions of a linear homogenous ODE on some interval, prove that $af+bg$ is also a solution (on that same
interval) for any real $a$ and $b$.

I found a solution online that essentially says:
Let, $y''+py'+qy=0$ be the linear homogenous ODE
Now check that $af+bg$ is a solution by verifying
$(af+bg)''+p(af+bg)'+q(af+bg)=af''+bg''+paf'+pbg'+qaf+qbg$
$=a(f''+pf'+qf)+b(g''+pg'+qg)=a*0+b*0$ (since $f$ and $g$ are solutions which means $f''+pf'+qf=0$ and $g''+pg'+qg=0$)
My question is, why did the solution use a second order ODE? Can I do this with a first order? i.e.:
$y'+py=0$ so $(af+bg)'+p(af+bg)=0$
$af'+bg'+paf+pbg=0$
$a(f'+pf)+b(g'+pg)=0$ Using the same reasoning as the solution above:
$a*0+b*0=0$

b) If we drop the "homogenous" hypothesis, is this still true? Prove a counterexample or proof.

My attempt at a solution
$y''+py'+qy=r(x)$
$f''+pf'+qf=0$ and $g''+pg'+qg=0$
Also the general solution to a non-homogenous DE is:
$y(x)=g(x)+h(x)$ where $h(x)$ is a particular solution and $g(x)$ is the general solution to the corresponding homogenous DE
$(af+bg)''+p(af+bg)'+q(af+bg)=a(f''+pf'+qf)+b(g''+pg'+qg)=a*0+b*0=0$
This is the general solution to the corresponding DE (i.e. the $g(x)$)
I'm stuck on how to proceed from here.
 A: For a) it is correct. Note we can also, define the diffential operator $L[y]=y''+p(t)+q(t)y=0$. Then if $f$ and $g$ they are solutions then $L[f]=0$ and $L[g]=0$. But then, $L[af+bg]=aL[f]+bL[g]=0$, then $af+bg$ is also one solutions for every value of the constants $a$ and $b$, that is basically the principle of superposition.
For your question it is a little harder to explain for me.  We can show that if we have one DE linear of form $y'=a(t)y$ then the vector space $E$ of solutions is isomorphic to ${\bf R}^1$, that is, we have one real solution. Indeed, fix $t_0$ in the interval of definition of the DE, then define the map $\phi_{t_0}(y)=y(t_0)$ for all $y\in E$ and we can show that $\phi_{t_0}$ is a linear transformation and with a little more effort (we need to invoke an existence and uniqueness theorem) we can guarantee injectivity and surjective and so $\phi_{t_0}$ is an isomorphim between the space of solutions $E$ and ${\bf R}^1$ and since $\dim {\bf R}^1=1$ so $\dim E=1$ as desired. But, also we can extend this result for DE linear of order $n$ and to show that it has $n$ solutions in the fundamental set of solutions. Returning with that fact for your question, noticed that if $f$ and $g$ they are solutions of $y'=a(t)y$ and we know that $E={\rm span}\{h(t)\}$ then $f(t)=ah(t)$ and $g(t)=bh(t)$ for some non zero constants $a$ and $b$. Thus $f=kg$ for some non zero constant $k$. But thus, they are linearly dependent. In less strict words: every time you think you have found a different solution for the DE, it can always be spanning by the fundamental set of solution $E$. Hence, one usually indicates this as obtaining a one-parametric family of solutions. This should be intuitive when one is familiar with the notion of a basis for a vector space.
For the second one literal, I do not understand why you suppose that $f$ and $g$ they are solutions of the homogeneous DE linear if in the problem says "drop homogeneous hypotheses". But if it is not homogeneous, then it must be inhomogeneous. So from what I understand you should suppose that $f$ and $g$ they are solutions for non-homogeneous DE linear and we want study $af+bg$ for that DE. If that is the case, you should suppose that $L[f]=r(t)$ and $L[g]=r(t)$ for $r(t)\not= 0$ and to answer: is $L[af+bg]=r(x)$? If the answer is not, you should consider an counterexample according to the order of the problem.
