Consider the linear homogeneous second order ODE
\[y''+p(x)y'+q(x)y=0. \;\;\;\;(31)\]
Principle of superposition: If \(y_1\) and \(y_2\) are solutions of (31),
then \(c_1y_1+c_2y_2\) is also a solution of (31). (verify)
Question. If \(y\) is a solution of (31), is it true that \(y=c_1y_1+c_2y_2\) for some scalars \(c_1\) and \(c_2\)?
Answer. Yes if \(y_1\) and \(y_2\) are linearly independent functions.
Definition.
\(y_1,y_2,\ldots,y_n\) are linearly independent functions if
\[c_1y_1+c_2y_2+\cdots+c_ny_n=0\implies c_1=c_2=\cdots=c_n=0.\]
Theorem.
\(y_1,y_2,\ldots,y_n\) are linearly dependent functions if and only if at least one of them is a linear combination of others.
Example.
\(y_1=x+1\) and \(y_2=x-1\) are linearly independent. Because if \(c_1y_1+c_2y_2=0\), then
\[\begin{align*}
&c_1(x+1)+c_2(x-1)=0\\
\implies& (c_1+c_2)x+(c_1-c_2)=0\\
\implies& c_1+c_2=0,\; c_1-c_2=0\\
\implies& c_1=c_2=0.
\end{align*}\]
\(y_1=e^x\) and \(y_2=e^{2x}\) are linearly independent.
\(y_1=x\) and \(y_2=2x\) are linearly dependent since \(-2y_1+y_2=0\).
The Wronskian of \(y_1\) and \(y_2\) is denoted by \(W(y_1,\,y_2)\) and defined by
\[W(y_1,\,y_2)=\begin{array}{|cc|}y_1&y_2\\y_1'&y_2'\end{array}=y_1y_2'-y_2y_1'.\]
Example.
Theorem.
\(y_1\) and \(y_2\) are linearly independent functions if and only if \(W(y_1,\,y_2)\neq 0\).
In other words, \(y_1\) and \(y_2\) are linearly dependent functions if and only if \(W(y_1,\,y_2)=0\).
\((\Longrightarrow)\) Assume that \(y_1\) and \(y_2\) are linearly dependent functions.
Then one of \(y_1\) and \(y_2\) is a multiple of the other, say \(y_2=cy_1\) for some scalar \(c\).
Then \[W(y_1,\,y_2)=W(y_1,\,cy_1)=y_1\cdot c_1y_1'-cy_1\cdot y_1'=0.\]
\((\Longleftarrow)\) Assume that \(W(y_1,\,y_2)=y_1y_2'-y_2y_1'=0\). Note that
\[\frac{d}{dx}\left(\frac{y_2}{y_1} \right)=\frac{y_1y_2'-y_2y_1'}{y_1^2}=0.\]
Then \(\frac{y_2}{y_1}=c,\) constant. Thus \(y_2=cy_1\) and consequently \(y_1\) and \(y_2\) are linearly dependent functions. \(\blacksquare\)
Theorem.
Suppose that \(y_1\) and \(y_2\) are solutions of the IVP
\[y''+p(x)y'+q(x)y=0,\;\; y(x_0)=r,\, y'(x_0)=s.\]
Then \(c_1y_1+c_2y_2\) is also a solution for some scalars \(c_1\) and \(c_2\) if and only if \(W(y_1,\,y_2)(x_0)\neq 0\).
\((\Longleftarrow)\) Assume that \(W(y_1,\,y_2)(x_0)\neq 0\). We will find \(c_1\) and \(c_2\) such that \(c_1y_1+c_2y_2\) is also a solution of the given IVP. First note that since \(y_1\) and \(y_2\) are solutions of \(y''+p(x)y'+q(x)y=0\), \(c_1y_1+c_2y_2\) is also a solution of \(y''+p(x)y'+q(x)y=0\) by the principle of superposition. Now we will find \(c_1\) and \(c_2\) such that \(c_1y_1+c_2y_2\) satisfies the initial conditions.
\[\begin{align*}
y(x_0)=r&\implies c_1y_1(x_0)+c_2y_2(x_0)=r\\
y(x_0)=s&\implies c_1y_1'(x_0)+c_2y_2'(x_0)=s
\end{align*}\]
Since \(W(y_1,\,y_2)(x_0)\neq 0\), solving for \(c_1\) and \(c_2\) we get
\[c_1=\frac{ry_2'(x_0)-sy_2(x_0)}{y_1(x_0)y_2'(x_0)-y_2(x_0)y_1'(x_0)}=\frac{ry_2'(x_0)-sy_2(x_0)}{W(y_1,\,y_2)(x_0)},\]
\[c_2=\frac{-ry_1'(x_0)+sy_1(x_0)}{y_1(x_0)y_2'(x_0)-y_2(x_0)y_1'(x_0)}=\frac{-ry_1'(x_0)+sy_1(x_0)}{W(y_1,\,y_2)(x_0)}.\]
\((\Longrightarrow)\) Similar. \(\blacksquare\)
Theorem.
Suppose that \(y_1\) and \(y_2\) are solutions of
\[y''+p(x)y'+q(x)y=0.\]
Then any solution \(y\) can be written as \(y=c_1y_1+c_2y_2\) for some scalars \(c_1\) and \(c_2\) if and only if \(W(y_1,\,y_2)(x_0)\neq 0\) for some \(x_0\).
\((\Longleftarrow)\) Assume \(W(y_1,\,y_2)(x_0)\neq 0\). Consider an arbitrary solution \(y=\phi(x)\) of the IVP
\[y''+p(x)y'+q(x)y=0,\; y(x_0)=\phi(x_0),y'(x_0)=\phi'(x_0).\]
By the preceding theorem \(y=c_1y_1+c_2y_2\) is a solution for some scalars \(c_1\) and \(c_2\).
Then \(\phi=y=c_1y_1+c_2y_2\) by the uniqueness of the solution of the preceding IVP.
\((\Longrightarrow)\) It follows from the linear independence of the fundamental solutions \(y_1\) and \(y_2\). \(\blacksquare\)
Note:
It can be shown for \(y_1\) and \(y_2\) in the preceding theorem that \(W(y_1,\,y_2)(x)=0\) either for all values of \(x\) or no values of \(x\). (see Abel's theorem)
When \(W(y_1,\,y_2)\neq 0\), \(y_1\) and \(y_2\) are linearly independent and any solution \(y\)
can be written as \(y=c_1y_1+c_2y_2\) for some scalars \(c_1\) and \(c_2\). That is why they are called
fundamental solutions of \(y''+p(x)y'+q(x)y=0\) and the general solution of
\(y''+p(x)y'+q(x)y=0\) is
\[y=c_1y_1+c_2y_2\]
for arbitrary scalars \(c_1\) and \(c_2\).
Example.
Recall from Section 3.1 that \(y''-2y'-3y=0\) has the general solution
\[y=c_1e^{-x}+c_2e^{3x}.\]
Here \(y_1=e^{-x}\) and \(y_2=e^{3x}\) are fundamental solutions because
\[W(e^{-x},\,e^{3x})=\,\begin{array}{|cc|}e^{-x}&e^{3x}\\(e^{-x})'&(e^{3x})'\end{array}
=\,\begin{array}{|cc|}e^{-x}&e^{3x}\\-e^{-x}&3e^{3x}\end{array}
=e^{-x}\cdot 3e^{3x}-e^{3x}\cdot(-e^{-x})=4e^{2x}\neq 0.\]
Example.
Consider \(y_1=e^{r_1x}\) and \(y_2=e^{r_2x}\), \(r_1\neq r_2\).
\[W(e^{r_1x},\,e^{r_2x})=\,\begin{array}{|cc|}e^{r_1x}&e^{r_2x}\\(e^{r_1x})'&(e^{r_2x})'\end{array}
=\,\begin{array}{|cc|}e^{r_1x}&e^{r_2x}\\r_1e^{r_1x}&r_2e^{r_2x}\end{array}
=e^{r_1x}\cdot r_2e^{r_2x}-e^{r_2x}\cdot r_1e^{r_1x})=(r_2-r_1)e^{(r_1+r_2)x}\neq 0.\]
For \(ay''+by'+cy=0\), if the roots \(r_1\) and \(r_2\) of the characteristic equation \(ar^2+br+c=0\) are real and distinct, then \(y=e^{r_1x}\) and \(y=e^{r_2x}\) are fundamental solutions and the general solution is
\[y=c_1e^{r_1x}+c_2e^{r_2x}.\]