Proof for general solution of homogeneous second-order linear ODE? In a differential equations class the professor stated that the general solution of a homogeneous second-order linear ODE would be in the form:
$$y = c_1y_1 + c_2y_2$$
Where $y_1$ and $y_2$ were distinct solutions of the ODE:
$$\frac{d^2y}{dx^2} + A\frac{dy}{dx} + By = 0$$
Where $A$ and $B$ are constant coefficients. I understand how to get the distinct solutions however what i don't get is why the general solution is in the form it is and what is the proof behind it? 
 A: Take a solution $y$, and for now assume that $y=c_1y_1+c_2y_2$. Consider $y(t_0)=y_0$ and $y'(t_0)=y_0'$. Thus we have
$$y(t_0)=c_1y_1(t_0)+c_2y_2(t_0)=y_0$$
$$\text{and}$$
$$y'(t_0)=c_1y_1'(t_0)+c_2y_2'(t_o)=y_0'$$
We now must solve for $c_1$ and $c_2$. We have the linear system
$$\left(\begin{array}{cc}
y_1(t_0) & y_2(t_0) \\
y_1'(t_0) & y_2'(t_0)
\end{array}\right)
\left(\begin{array}{c}
c_1 \\
c_2
\end{array}\right)=
\left(\begin{array}{c}
y_0 \\
y_0'
\end{array}\right)\qquad (1)$$
Now since $y_1$ and $y_2$ are fundamental solutions, meaning their Wronskian
$$W[y_1,y_2]=\left|
\begin{array}{cc}
y_1(t) & y_2(t) \\
y_1'(t) & y_2'(t)
\end{array}
\right|\neq 0$$
$\forall t$ in the interval upon which $y_1$ and $y_2$ are being taken as solutions we have the explicit solution to $(1)$ which is
$$\left(\begin{array}{c}
c_1 \\
c_2
\end{array}\right)=\frac{1}{W[y_1,y_2](t_0)}\left(\begin{array}{cc}
y_2'(t_0) & -y_2(t_0) \\
-y_1'(t_0) & y_1(t_0)
\end{array}\right)\left(\begin{array}{c}
y_0 \\
y_0'
\end{array}\right)$$
Hence there exists a $c_1$ and $c_2$ such that $y=c_1y_1+c_2y_2$ and by the Existence-Uniqueness Theorem this is the only possible $y$ such that $y(t_0)=y_0$ and $y'(t_0)=y_0'$, forcing $y$ to be of this form. QED

Above I gave the proof for why all solutions to the homogeneous linear ODE must to be of the form $y=c_1y_1+c_2y_2$. But if you only wish to convince yourself why $y=c_1y_1+c_2y_2$ is a solution then realize that since the operator $L[y]$,
$$L[y]=y''+Ay'+By$$
is linear we have
$$L[c_1y_1+c_2y_2]=L[c_1y_1]+L[c_2y_2]=c_1L[y_1]+c_2L[y_2]=0$$
because, since $y_1$ and $y_2$ are solutions to the DE, $L[y_1]=L[y_2]=0$ and since $L[c_1y_1+c_2y_2]=0$ it is also a solution to the DE.
A: Another option is to explicitly construct the general solution to the differential equation
$$y''+Ay'+By=0,$$
where $y=y(x)$.
This can be done by transforming it to a first order linear differential equation, for which we know how to find the general solution.
If $B=0$ we can simply integrate both sides to get a first order differential equation (with general solution $y=c_1+c_2e^{-Ax}$), so we may henceforth assume that $B \neq 0$.
Now let $z=z(x)$. A change of variables $y=ze^{rx}$ (for any constant $r\neq0$) makes $y'=(z'+rz)e^{rx}$ and $y''=(z''+2rz'+r^2z)e^{rx}$. Insertion into the original equation yields
$$
y''+Ay'+By
= (z''+2rz'+r^2z)e^{rx}+A(z'+rz)e^{rx}+Bze^{rx}
= e^{rx}(z''+(2r+A)z'+(r^2+Ar+B)z)=0 \Longleftrightarrow \\
z''+(2r+A)z'+(r^2+Ar+B)z=0.
$$
Now we choose $r$ such that $r^2+Ar+B=0$. Here we must consider two cases:

*

*The equation $r^2+Ar+B=0$ has a double root $r$. In this case $z''=0 \Rightarrow z=c_1x+c_2$, so $y=ze^{rx}=(c_1x+c_2)e^{rx}$.

*The equation $r^2+Ar+B=0$ has distinct roots $r_1$ and $r_2$ (assume we use $r=r_1$). In this case
$$
z''+(2r_1+A)z' = z''+(r_1-r_2)z'=0 \Longrightarrow z'+(r_1-r_2)z=M,
$$
which has the general solution $z=\frac{M}{r_1-r2}+c_2e^{(r_2-r_1)x}=c_1+c_2e^{(r_2-r_1)x}$. This means that $y=c_1e^{r_1x}+c_2e^{r_2x}$.

We have now exhausted the possibilities, and in each case the general solution is of the form $y=c_1y_1+c_2y_2$, as desired.
