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2.1: Linear Second Order Homogeneous Equations

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

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A second order differential equation is said to be linear if it can be written as

\label{eq:2.1.1}
y''+p(x)y'+q(x)y=f(x).

We call the function $$f$$ on the right a forcing function, since in physical applications it's often related to a force acting on some system modeled by the differential equation. We say that \eqref{eq:2.1.1} is $$\textcolor{blue}{\mbox{homogeneous}}$$ if $$f \equiv 0$$ or $$\textcolor{blue}{\mbox{nonhomogeneous}}$$ if $$f\not\equiv 0$$. Since these definitions are like the corresponding definitions in 3.3: First order linear equations for the linear first order equation

\label{eq:2.1.2}
y'+p(x)y=f(x),

it's natural to expect similarities between methods of solving \eqref{eq:2.1.1} and \eqref{eq:2.1.2}. However, solving \eqref{eq:2.1.1} is more difficult than solving \eqref{eq:2.1.2}. For example, while Theorem $$(2.1.1)$$ gives a formula for the general solution of \eqref{eq:2.1.2} in the case where $$f\equiv0$$ and Theorem 2.2.2 gives a formula for the case where $$f\not\equiv0$$, there are no formulas for the general solution of \eqref{eq:2.1.1} in either case. Therefore we must be content to solve linear second order equations of special forms.

In Section 2.1 we considered the homogeneous equation $$y'+p(x)y=0$$ first, and then used a nontrivial solution of this equation to find the general solution of the nonhomogeneous equation $$y'+p(x)y=f(x)$$. Although the progression from the homogeneous to the nonhomogeneous case isn't that simple for the linear second order
equation, it's still necessary to solve the homogeneous equation

\label{eq:2.1.3}
y''+p(x)y'+q(x)y=0

in order to solve the nonhomogeneous equation \eqref{eq:2.1.1}. This section is devoted to \eqref{eq:2.1.3}.

The next theorem gives sufficient conditions for existence and uniqueness of solutions of initial value problems for \eqref{eq:5.1.3}. We omit the proof.

Theorem $$\PageIndex{1}$$

Suppose $$p$$ and $$q$$ are continuous on an open interval $$(a,b),$$ let $$x_0$$ be any point in $$(a,b),$$ and let $$k_0$$ and $$k_1$$ be arbitrary real numbers$$.$$ Then the initial value problem

\begin{eqnarray*}
y''+p(x)y'+q(x)y=0,\ y(x_0)=k_0,\ y'(x_0)=k_1
\end{eqnarray*}

has a unique solution on $$(a,b).$$

Proof

Since $$y\equiv0$$ is obviously a solution of \eqref{eq:2.1.3} we call it the $$\textcolor{blue}{\mbox{trivial}}$$ solution. Any other solution is $$\textcolor{blue}{\mbox{nontrivial}}$$. Under the assumptions of Theorem $$(2.1.1)$$, the only solution of the initial value problem

\begin{eqnarray*}
y''+p(x)y'+q(x)y=0,\ y(x_0)=0,\ y'(x_0)=0
\end{eqnarray*}

on $$(a,b)$$ is the trivial solution. (Exercise $$(2.1E.24)$$).

The next three examples illustrate concepts that we'll develop later in this section. You shouldn't be concerned with how to $$\textcolor{blue}{\mbox{find}}$$ the given solutions of the equations in these examples. This will be explained in later sections.

Example $$\PageIndex{1}$$

The coefficients of $$y'$$ and $$y$$ in

\label{eq:2.1.4}
y''-y=0

are the constant functions $$p\equiv0$$ and $$q\equiv-1$$, which are continuous on $$(-\infty,\infty)$$. Therefore Theorem $$(2.1.1)$$ implies that every initial value problem for \eqref{eq:2.1.4} has a unique solution on $$(-\infty,\infty)$$.

(a) Verify that $$y_1=e^x$$ and $$y_2=e^{-x}$$ are solutions of \eqref{eq:2.1.4} on $$(-\infty,\infty)$$.

(b) Verify that if $$c_1$$ and $$c_2$$ are arbitrary constants, $$y=c_1e^x+c_2e^{-x}$$ is a solution of \eqref{eq:2.1.4} on $$(-\infty,\infty)$$.

(c) Solve the initial value problem

\label{eq:2.1.5}

(a) If $$y_1=e^x$$ then $$y_1'=e^x$$ and $$y_1''=e^x=y_1$$, so $$y_1''-y_1=0$$. If $$y_2=e^{-x}$$, then $$y_2'=-e^{-x}$$ and $$y_2''=e^{-x}=y_2$$, so $$y_2''-y_2=0$$.

(b) If

\label{eq:2.1.6}
y=c_1e^x+c_2e^{-x}

then

\label{eq:2.1.7}
y'=c_1e^x-c_2e^{-x}

and

\begin{eqnarray*}
y''=c_1e^x+c_2e^{-x},
\end{eqnarray*}

so

\begin{eqnarray*}
y''-y&=&(c_1e^x+c_2e^{-x})-(c_1e^x+c_2e^{-x})\\
&=&c_1(e^x-e^x)+c_2(e^{-x}-e^{-x})=0
\end{eqnarray*}

for all $$x$$. Therefore $$y=c_1e^x+c_2e^{-x}$$ is a solution of \eqref{eq:2.1.4} on $$(-\infty,\infty)$$.

(c) We can solve \eqref{eq:2.1.5} by choosing $$c_1$$ and $$c_2$$ in \eqref{eq:2.1.6} so that $$y(0)=1$$ and $$y'(0)=3$$. Setting $$x=0$$ in \eqref{eq:2.1.6} and \eqref{eq:2.1.7} shows that this is equivalent to

\begin{eqnarray*}
c_1+c_2&=&1\\
c_1-c_2&=&3.
\end{eqnarray*}

Solving these equations yields $$c_1=2$$ and $$c_2=-1$$. Therefore $$y=2e^x-e^{-x}$$ is the unique solution of \eqref{eq:2.1.5} on $$(-\infty,\infty)$$.

Example $$\PageIndex{2}$$

Let $$\omega$$ be a positive constant. The coefficients of $$y'$$ and $$y$$ in

\label{eq:2.1.8}
y''+\omega^2y=0

are the constant functions $$p\equiv0$$ and $$q\equiv\omega^2$$, which are continuous on $$(-\infty,\infty)$$. Therefore Theorem $$(2.1.1)$$ implies that every initial value problem for \eqref{eq:2.1.8} has a unique solution on $$(-\infty,\infty)$$.

(a) Verify that $$y_1=\cos\omega x$$ and $$y_2=\sin\omega x$$ are solutions of \eqref{eq:2.1.8} on $$(-\infty,\infty)$$.

(b) Verify that if $$c_1$$ and $$c_2$$ are arbitrary constants then $$y=c_1\cos\omega x+c_2\sin\omega x$$ is a solution of \eqref{eq:2.1.8} on $$(-\infty,\infty)$$.

(c) Solve the initial value problem

\label{eq:2.1.9}

(a) If $$y_1=\cos\omega x$$ then $$y_1'=-\omega\sin\omega x$$ and $$y_1''=-\omega^2\cos\omega x=-\omega^2y_1$$, so $$y_1''+\omega^2y_1=0$$. If $$y_2=\sin\omega x$$ then, $$y_2'=\omega\cos\omega x$$ and $$y_2''=-\omega^2\sin\omega x=-\omega^2y_2$$, so $$y_2''+\omega^2y_2=0$$.

(b) If

\label{eq:2.1.10}
y=c_1\cos\omega x+c_2\sin\omega x

then

\label{eq:2.1.11}
y'=\omega(-c_1\sin\omega x+c_2\cos\omega x)

and

\begin{eqnarray*}
y''=-\omega^2(c_1\cos\omega x+c_2\sin\omega x),
\end{eqnarray*}

so

\begin{eqnarray*}
y''+\omega^2y&=& -\omega^2(c_1\cos\omega x+c_2\sin\omega x)
+\omega^2(c_1\cos\omega x+c_2\sin\omega x)\\
&=&c_1\omega^2(-\cos\omega x+\cos\omega x)+
c_2\omega^2(-\sin\omega x+\sin\omega x)=0
\end{eqnarray*}

for all $$x$$. Therefore $$y=c_1\cos\omega x+c_2\sin\omega x$$ is a solution of \eqref{eq:2.1.8} on $$(-\infty,\infty)$$.

(c) To solve \eqref{eq:2.1.9}, we must choosing $$c_1$$ and $$c_2$$ in \eqref{eq:2.1.10} so that $$y(0)=1$$ and $$y'(0)=3$$. Setting $$x=0$$ in \eqref{eq:2.1.10} and \eqref{eq:2.1.11} shows that $$c_1=1$$ and $$c_2=3/\omega$$. Therefore

\begin{eqnarray*}
y=\cos\omega x+{3\over\omega}\sin\omega x
\end{eqnarray*}

is the unique solution of \eqref{eq:2.1.9} on $$(-\infty,\infty)$$.

Theorem $$(2.1.1)$$ implies that if $$k_0$$ and $$k_1$$ are arbitrary real numbers then the initial value problem

\label{eq:2.1.12}

has a unique solution on an interval $$(a,b)$$ that contains $$x_0$$, provided that $$P_0$$, $$P_1$$, and $$P_2$$ are continuous and $$P_0$$ has no zeros on $$(a,b)$$. To see this, we rewrite the differential equation in \eqref{eq:2.1.12} as

\begin{eqnarray*}
y''+{P_1(x)\over P_0(x)}y'+{P_2(x)\over P_0(x)}y=0
\end{eqnarray*}

and apply Theorem $$(2.1.1)$$ with $$p=P_1/P_0$$ and $$q=P_2/P_0$$.

Example $$\PageIndex{3}$$

The equation

\label{eq:2.1.13}
x^2y''+xy'-4y=0

has the form of the differential equation in \eqref{eq:2.1.12}, with $$P_0(x)=x^2$$, $$P_1(x)=x$$, and $$P_2(x)=-4$$, which are are all continuous on $$(-\infty,\infty)$$. However, since $$P(0)=0$$ we must consider solutions of \eqref{eq:2.1.13} on $$(-\infty,0)$$ and $$(0,\infty)$$. Since $$P_0$$ has no zeros on these intervals, Theorem $$(2.1.1)$$ implies that the initial value problem

\begin{eqnarray*}
\end{eqnarray*}

has a unique solution on $$(0,\infty)$$ if $$x_0>0$$, or on $$(-\infty,0)$$ if $$x_0<0$$.

(a) Verify that $$y_1=x^2$$ is a solution of \eqref{eq:2.1.13} on $$(-\infty,\infty)$$ and $$y_2=1/x^2$$ is a solution of \eqref{eq:2.1.13} on $$(-\infty,0)$$ and $$(0,\infty)$$.

(b) Verify that if $$c_1$$ and $$c_2$$ are any constants then $$y=c_1x^2+c_2/x^2$$ is a solution of \eqref{eq:2.1.13} on $$(-\infty,0)$$ and $$(0,\infty)$$.

(c) Solve the initial value problem

\label{eq:2.1.14}

(d) Solve the initial value problem

\label{eq:2.1.15}

(a) If $$y_1=x^2$$ then $$y_1'=2x$$ and $$y_1''=2$$, so

\begin{eqnarray*}
x^2y_1''+xy_1'-4y_1=x^2(2)+x(2x)-4x^2=0
\end{eqnarray*}

for $$x$$ in $$(-\infty,\infty)$$. If $$y_2=1/x^2$$, then $$y_2'=-2/x^3$$ and $$y_2''=6/x^4$$, so

\begin{eqnarray*}
x^2y_2''+xy_2'-4y_2=x^2\left(6\over x^4\right)-x\left(2\over x^3\right)-{4\over x^2}=0
\end{eqnarray*}

for $$x$$ in $$(-\infty,0)$$ or $$(0,\infty)$$.

(b) If

\label{eq:2.1.16}
y=c_1x^2+{c_2\over x^2}

then

\label{eq:2.1.17}
y'=2c_1x-{2c_2\over x^3}

and

\begin{eqnarray*}
y''=2c_1+{6c_2\over x^4},
\end{eqnarray*}

so

\begin{eqnarray*}
x^2y''+xy'-4y&=&x^2\displaystyle{\left(2c_1+{6c_2\over x^4}\right)} +x\displaystyle{\left(2c_1x-{2c_2\over x^3}\right)} -4\displaystyle{\left(c_1x^2+{c_2\over x^2}\right)}\\
&=&c_1(2x^2+2x^2-4x^2) +c_2\displaystyle{\left({6\over x^2}-{2\over x^2}-{4\over x^2}\right)}\\
&=&c_1\cdot0+c_2\cdot0=0
\end{eqnarray*}

for $$x$$ in $$(-\infty,0)$$ or $$(0,\infty)$$.

(c) To solve \eqref{eq:2.1.14}, we choose $$c_1$$ and $$c_2$$ in \eqref{eq:2.1.16} so that $$y(1)=2$$ and $$y'(1)=0$$. Setting $$x=1$$ in \eqref{eq:2.1.16} and \eqref{eq:2.1.17} shows that this is equivalent to

\begin{eqnarray*}
\phantom{2}c_1+\phantom{2}c_2&=&2\\
2c_1-2c_2&=&0.
\end{eqnarray*}

Solving these equations yields $$c_1=1$$ and $$c_2=1$$. Therefore $$y=x^2+1/x^2$$ is the unique solution of \eqref{eq:2.1.14} on $$(0,\infty)$$.

(d) We can solve \eqref{eq:2.1.15} by choosing $$c_1$$ and $$c_2$$ in \eqref{eq:2.1.16} so that $$y(-1)=2$$ and $$y'(-1)=0$$. Setting $$x=-1$$ in \eqref{eq:2.1.16} and \eqref{eq:2.1.17} shows that this is equivalent to

\begin{eqnarray*}
\phantom{-2}c_1+\phantom{2}c_2&=&2\\
-2c_1+2c_2&=&0.
\end{eqnarray*}

Solving these equations yields $$c_1=1$$ and $$c_2=1$$. Therefore $$y=x^2+1/x^2$$ is the unique solution of \eqref{eq:2.1.15} on $$(-\infty,0)$$.

Although the $$\textcolor{blue}{\mbox{formulas}}$$ for the solutions of \eqref{eq:2.1.14} and \eqref{eq:2.1.15} are both $$y=x^2+1/x^2$$, you should not conclude that these two initial value problems have the same solution. Remember that a solution of an initial value problem is defined $$\textcolor{blue}{\mbox{on an interval that contains the initial point}}$$; therefore, the solution of \eqref{eq:2.1.14} is $$y=x^2+1/x^2$$ $$\textcolor{blue}{\mbox{on the interval}}$$ $$(0,\infty)$$, which contains the initial point $$x_0=1$$, while the solution of \eqref{eq:2.1.15} is $$y=x^2+1/x^2$$ $$\textcolor{blue}{\mbox{on the interval}}$$ $$(-\infty,0)$$, which contains the initial point $$x_0=-1$$.

The General Solution of a Homogeneous Linear Second Order Equation

If $$y_1$$ and $$y_2$$ are defined on an interval $$(a,b)$$ and $$c_1$$ and $$c_2$$ are constants, then

\begin{eqnarray*}
y=c_1y_1+c_2y_2
\end{eqnarray*}

is a $$\textcolor{blue}{\mbox{linear combination of \(y_1$$ and $$y_2$$}} \). For example, $$y=2\cos x+7 \sin x$$ is a linear combination of $$y_1= \cos x$$ and $$y_2=\sin x$$, with $$c_1=2$$ and $$c_2=7$$.

The next theorem states a fact that we've already verified in Examples $$(2.1.1)$$, $$(2.1.2)$$, and $$(2.1.3)$$.

Theorem $$\PageIndex{2}$$

If $$y_1$$ and $$y_2$$ are solutions of the homogeneous equation

\label{eq:2.1.18}
y''+p(x)y'+q(x)y=0

on $$(a,b),$$ then any linear combination

\label{eq:2.1.19}
y=c_1y_1+c_2y_2

of $$y_1$$ and $$y_2$$ is also a solution of \eqref{eq:2.1.18} on $$(a,b).$$

Proof

If

\begin{eqnarray*}
y=c_1y_1+c_2y_2
\end{eqnarray*}

then

\begin{eqnarray*}
y'=c_1y_1'+c_2y_2'\mbox{ and } y''=c_1y_1''+c_2y_2''.
\end{eqnarray*}

Therefore

\begin{eqnarray*}
y''+p(x)y'+q(x)y&=&(c_1y_1''+c_2y_2'')+p(x)(c_1y_1'+c_2y_2') +q(x)(c_1y_1+c_2y_2)\\
&=&c_1\left(y_1''+p(x)y_1'+q(x)y_1\right) +c_2\left(y_2''+p(x)y_2'+q(x)y_2\right)\\
&=&c_1\cdot0+c_2\cdot0=0,
\end{eqnarray*}

since $$y_1$$ and $$y_2$$ are solutions of \eqref{eq:2.1.18}.

We say that $$\{y_1,y_2\}$$ is a $$\textcolor{blue}{\mbox{fundamental set of solutions of \(\eqref{eq:2.1.18}$$ on}} \) $$(a,b)$$ if every solution of \eqref{eq:2.1.18} on $$(a,b)$$ can be written as a linear combination of $$y_1$$ and $$y_2$$ as in \eqref{eq:2.1.19}. In this case we say that \eqref{eq:2.1.19} is $$\textcolor{blue}{\mbox{general solution of \(\eqref{eq:2.1.18}$$ on}} \) $$(a,b)$$.

Linear Independence

We need a way to determine whether a given set $$\{y_1,y_2\}$$ of solutions of \eqref{eq:2.1.18} is a fundamental set. The next definition will enable us to state necessary and
sufficient conditions for this.

We say that two functions $$y_1$$ and $$y_2$$ defined on an interval $$(a,b)$$ are $$\textcolor{blue}{\mbox{linearly independent on}}$$ $$(a,b)$$ if neither is a constant multiple of the other on $$(a,b)$$. (In particular, this means that neither can be the trivial solution of \eqref{eq:2.1.18}, since, for example, if $$y_1\equiv0$$ we could write $$y_1=0y_2$$.) We'll also say that the set $$\{y_1,y_2\}$$ $$\textcolor{blue}{\mbox{is linearly independent on}}$$ $$(a,b)$$.

Theorem $$\PageIndex{3}$$

Suppose $$p$$ and $$q$$ are continuous on $$(a,b).$$ Then a set $$\{y_1,y_2\}$$ of solutions of

\label{eq:2.1.20}
y''+p(x)y'+q(x)y=0

on $$(a,b)$$ is a fundamental set if and only if $$\{y_1,y_2\}$$ is linearly independent on $$(a,b).$$

Proof

Add proof here and it will automatically be hidden if you have a "AutoNum" template active on the page.

We'll present the proof of Theorem $$(2.1.3)$$ in steps worth regarding as theorems in their own right. However, let's first interpret Theorem $$(2.1.3)$$ in terms of Examples $$(2.1.1)$$, $$(2.1.2)$$, and $$(2.1.3)$$.

Example $$\PageIndex{4}$$:

(a) Since $$e^x/e^{-x}=e^{2x}$$ is nonconstant, Theorem $$(2.1.3)$$ implies that $$y=c_1e^x+c_2e^{-x}$$ is the general solution of $$y''-y=0$$ on $$(-\infty,\infty)$$.

(b) Since $$\cos\omega x/\sin\omega x=\cot\omega x$$ is nonconstant, Theorem $$(2.1.3)$$ implies that $$y=c_1\cos\omega x+c_2\sin\omega x$$ is the general solution
of $$y''+\omega^2y=0$$ on $$(-\infty,\infty)$$.

(c) Since $$x^2/x^{-2}=x^4$$ is nonconstant, Theorem $$(2.1.3)$$ implies that $$y=c_1x^2+c_2/x^2$$ is the general solution of $$x^2y''+xy'-4y=0$$ on $$(-\infty,0)$$ and $$(0,\infty)$$.

The Wronskian and Abel's Formula

To motivate a result that we need in order to prove Theorem $$(2.1.3)$$, let's see what is required to prove that$$\{y_1,y_2\}$$ is a fundamental set of solutions of \eqref{eq:2.1.20} on $$(a,b)$$. Let $$x_0$$ be an arbitrary point in $$(a,b)$$, and suppose $$y$$ is an arbitrary solution of \eqref{eq:2.1.20} on $$(a,b)$$. Then $$y$$ is the unique solution of the initial value problem

\label{eq:2.1.21}

that is, $$k_0$$ and $$k_1$$ are the numbers obtained by evaluating $$y$$ and $$y'$$ at $$x_0$$. Moreover, $$k_0$$ and $$k_1$$ can be any real numbers, since Theorem $$(2.1.1)$$ implies that \eqref{eq:2.1.21} has a solution no matter how $$k_0$$ and $$k_1$$ are chosen. Therefore $$\{y_1,y_2\}$$ is a fundamental set of solutions of \eqref{eq:2.1.20} on $$(a,b)$$ if and only if it's possible to write the solution of an arbitrary initial value problem \eqref{eq:2.1.21} as $$y=c_1y_1+c_2y_2$$. This is equivalent to requiring that the system

\label{eq:2.1.22}
\begin{array}{rcl}
c_1y_1(x_0)+c_2y_2(x_0)&=&k_0\\
c_1y_1'(x_0)+c_2y_2'(x_0)&=&k_1
\end{array}

has a solution $$(c_1,c_2)$$ for every choice of $$(k_0,k_1)$$. Let's try to solve \eqref{eq:2.1.22}.

Multiplying the first equation in \eqref{eq:2.1.22} by $$y_2'(x_0)$$ and the second by $$y_2(x_0)$$ yields

\begin{eqnarray*}
c_1y_1(x_0)y_2'(x_0)+c_2y_2(x_0)y_2'(x_0)&=& y_2'(x_0)k_0\\
c_1y_1'(x_0)y_2(x_0)+c_2y_2'(x_0)y_2(x_0)&=& y_2(x_0)k_1,
\end{eqnarray*}

and subtracting the second equation here from the first yields

\label{eq:2.1.23}
\left(y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0)\right)c_1= y_2'(x_0)k_0-y_2(x_0)k_1.

Multiplying the first equation in \eqref{eq:2.1.22} by $$y_1'(x_0)$$ and the second by $$y_1(x_0)$$ yields

\begin{eqnarray*}
c_1y_1(x_0)y_1'(x_0)+c_2y_2(x_0)y_1'(x_0)&=& y_1'(x_0)k_0\\
c_1y_1'(x_0)y_1(x_0)+c_2y_2'(x_0)y_1(x_0)&=& y_1(x_0)k_1,
\end{eqnarray*}

and subtracting the first equation here from the second yields

\label{eq:2.1.24}
\left(y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0)\right)c_2=y_1(x_0)k_1-y_1'(x_0)k_0.

If

\begin{eqnarray*}
y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0)=0,
\end{eqnarray*}

it's impossible to satisfy \eqref{eq:2.1.23} and \eqref{eq:2.1.24} (and therefore \eqref{eq:2.1.22}) unless $$k_0$$ and $$k_1$$ happen to satisfy

\begin{eqnarray*}
y_1(x_0)k_1-y_1'(x_0)k_0&=&0\\
y_2'(x_0)k_0-y_2(x_0)k_1&=&0.
\end{eqnarray*}

On the other hand, if

\label{eq:2.1.25}
y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0)\ne0

we can divide \eqref{eq:2.1.23} and \eqref{eq:2.1.24} through by the quantity on the left to obtain

\label{eq:2.1.26}
\begin{array}{rcl}
c_1&=&\displaystyle{y_2'(x_0)k_0-y_2(x_0)k_1\over y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0)}\\
c_2&=&\displaystyle{y_1(x_0)k_1-y_1'(x_0)k_0\over y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0)},
\end{array}

no matter how $$k_0$$ and $$k_1$$ are chosen. This motivates us to consider conditions on $$y_1$$ and $$y_2$$ that imply \eqref{eq:2.1.25}.

Theorem $$\PageIndex{4}$$

Suppose $$p$$ and $$q$$ are continuous on $$(a,b),$$ let $$y_1$$ and $$y_2$$ be solutions of

\label{eq:2.1.27}
y''+p(x)y'+q(x)y=0

on $$(a,b)$$, and define

\label{eq:2.1.28}
W=y_1y_2'-y_1'y_2.

Let $$x_0$$ be any point in $$(a,b).$$ Then

\label{eq:2.1.29}

Therefore either $$W$$ has no zeros in $$(a,b)$$ or $$W\equiv0$$ on $$(a,b).$$

Proof

Differentiating \eqref{eq:2.1.28} yields

\label{eq:2.1.30}
W'=y'_1y'_2+y_1y''_2-y'_1y'_2-y''_1y_2=y_1y''_2-y''_1y_2.

Since $$y_1$$ and $$y_2$$ both satisfy \eqref{eq:2.1.27},

\begin{eqnarray*}
y''_1 =-py'_1-qy_1\mbox{ and } y''_2 =-py'_2-qy_2.
\end{eqnarray*}

Substituting these into \eqref{eq:2.1.30} yields

\begin{eqnarray*}
W'&=& \displaystyle -y_1\bigl(py'_2+qy_2\bigr) +y_2\bigl(py'_1+qy_1\bigr) \\
&=& \displaystyle -p(y_1y'_2-y_2y'_1)-q(y_1y_2-y_2y_1)\\
&=& -p(y_1y'_2-y_2y'_1)=-pW.
\end{eqnarray*}

Therefore $$W'+p(x)W=0$$; that is, $$W$$ is the solution of the initial value problem

\begin{eqnarray*}
\end{eqnarray*}

We leave it to you to verify by separation of variables that this implies \eqref{eq:2.1.29}. If $$W(x_0)\ne0$$, \eqref{eq:2.1.29} implies that $$W$$ has no zeros in $$(a,b)$$, since an exponential is never zero. On the other hand, if $$W(x_0)=0$$, \eqref{eq:2.1.29} implies that $$W(x)=0$$ for all $$x$$ in $$(a,b)$$.

The function $$W$$ defined in \eqref{eq:2.1.28} is the Wronskian of $$\{y_1,y_2\}$$. Formula \eqref{eq:2.1.29} is Abel's formula.

The Wronskian of $$\{y_1,y_2\}$$ is usually written as the determinant

\begin{eqnarray*}
W=\left| \begin{array}{cc}
y_1 & y_2 \\
y'_1 & y'_2
\end{array} \right|.
\end{eqnarray*}

The expressions in \eqref{eq:2.1.26} for $$c_1$$ and $$c_2$$ can be written in terms of determinants as

\begin{eqnarray*}
c_1={1\over W(x_0)}
\left| \begin{array}{cc}
k_0 & y_2(x_0) \\
k_1 & y'_2(x_0)
\end{array} \right|
\mbox{ and }
c_2={1\over W(x_0)}
\left| \begin{array}{cc}
y_1(x_0) & k_0 \\
y'_1(x_0) &k_1
\end{array} \right|.
\end{eqnarray*}

If you've taken linear algebra you may recognize this as Cramer's rule.

Example $$\PageIndex{5}$$

Verify Abel's formula for the following differential equations and the corresponding solutions, from Examples $$(2.1.1)$$. $$(2.1.2)$$, and $$(2.1.3)$$:

(a) $$y''-y=0;\quad y_1=e^x,\; y_2=e^{-x}$$

(b) $$y''+\omega^2y=0;\quad \quad y_1=\cos\omega x,\; y_2=\sin\omega x$$

(c) $$x^2y''+xy'-4y=0;\quad y_1=x^2,\; y_2=1/x^2$$

(a) Since $$p\equiv0$$, we can verify Abel's formula by showing that $$W$$ is constant, which is true, since

\begin{eqnarray*}
W(x)=\left| \begin{array}{rr}
e^x & e^{-x} \\
e^x & -e^{-x}
\end{array} \right|=e^x(-e^{-x})-e^xe^{-x}=-2
\end{eqnarray*}

for all $$x$$.

(b) Again, since $$p\equiv0$$, we can verify Abel's formula by showing that $$W$$ is constant, which is true, since

\begin{eqnarray*}
W(x)&=&\displaystyle{\left| \begin{array}{cc}\cos\omega x & \sin\omega x \\
-\omega\sin\omega x &\omega\cos\omega x
\end{array} \right|}\\
&=&\cos\omega x (\omega\cos\omega x)-(-\omega\sin\omega x)\sin\omega x\\ &=&\omega(\cos^2\omega x+\sin^2\omega x)=\omega
\end{eqnarray*}

for all $$x$$.

(c) Computing the Wronskian of $$y_1=x^2$$ and $$y_2=1/x^2$$ directly yields

\label{eq:2.1.31}
W=\left| \begin{array}{cc}
x^2 & 1/x^2 \\
2x & -2/x^3
\end{array} \right|=x^2\left(-{2\over
x^3}\right)-2x\left(1\over x^2\right)=-{4\over x}.

To verify Abel's formula we rewrite the differential equation as

\begin{eqnarray*}
y''+{1\over x}y'-{4\over x^2}y=0
\end{eqnarray*}

to see that $$p(x)=1/x$$. If $$x_0$$ and $$x$$ are either both in $$(-\infty,0)$$ or both in $$(0,\infty)$$ then

\begin{eqnarray*}
\int_{x_0}^x p(t)\,dt=\int_{x_0}^x {dt\over t}=\ln\left(x\over x_0\right),
\end{eqnarray*}

so Abel's formula becomes

\begin{eqnarray*}
W(x)&=&W(x_0)e^{-\ln(x/x_0)}=W(x_0){x_0\over x}\\
&=&-\left(4\over x_0\right)\left(x_0\over x\right)\mbox{ from \eqref{eq:2.1.31}}\\
&=&-{4\over x},
\end{eqnarray*}

which is consistent with \eqref{eq:2.1.31}.

The next theorem will enable us to complete the proof of Theorem $$(2.1.3)$$.

Theorem $$\PageIndex{5}$$

Suppose $$p$$ and $$q$$ are continuous on an open interval $$(a,b),$$ let $$y_1$$ and $$y_2$$ be solutions of

\label{eq:2.1.32}
y''+p(x)y'+q(x)y=0

on $$(a,b),$$ and let $$W=y_1y_2'-y_1'y_2.$$ Then $$y_1$$ and $$y_2$$ are linearly independent on $$(a,b)$$ if and only if $$W$$ has no zeros on $$(a,b).$$

Proof

We first show that if $$W(x_0)=0$$ for some $$x_0$$ in $$(a,b)$$, then $$y_1$$ and $$y_2$$ are linearly dependent on $$(a,b)$$. Let $$I$$ be a subinterval of $$(a,b)$$ on which $$y_1$$ has no zeros. (If there's no such subinterval, $$y_1\equiv0$$ on $$(a,b)$$, so $$y_1$$ and $$y_2$$ are linearly independent, and we're finished with this part of the proof.) Then $$y_2/y_1$$ is defined on $$I$$, and

\label{eq:2.1.33}
\left(y_2\over y_1\right)'={y_1y_2'-y_1'y_2\over y_1^2}={W\over y_1^2}.

However, if $$W(x_0)=0$$, Theorem $$(2.1.4)$$ implies that $$W\equiv0$$ on $$(a,b)$$. Therefore \eqref{eq:2.1.33} implies that $$(y_2/y_1)'\equiv0$$, so $$y_2/y_1=c$$ (constant) on $$I$$. This shows that $$y_2(x)=cy_1(x)$$ for all $$x$$ in $$I$$. However, we want to show that $$y_2=cy_1(x)$$ for all $$x$$ in $$(a,b)$$. Let $$Y=y_2-cy_1$$. Then $$Y$$ is a solution of \eqref{eq:2.1.32} on $$(a,b)$$ such that $$Y\equiv0$$ on $$I$$, and therefore $$Y'\equiv0$$ on $$I$$. Consequently, if $$x_0$$ is chosen arbitrarily in $$I$$ then $$Y$$ is a solution of the initial value problem

\begin{eqnarray*}
\end{eqnarray*}

which implies that $$Y\equiv0$$ on $$(a,b)$$, by the paragraph following Theorem $$(2.1.1)$$. (See also Exercise $$(2.1E.24)$$. Hence, $$y_2-cy_1\equiv0$$
on $$(a,b)$$, which implies that $$y_1$$ and $$y_2$$ are not linearly independent on $$(a,b)$$.

Now suppose $$W$$ has no zeros on $$(a,b)$$. Then $$y_1$$ can't be identically zero on $$(a,b)$$ (why not?), and therefore there is a subinterval $$I$$ of $$(a,b)$$ on which $$y_1$$ has no zeros. Since \eqref{eq:2.1.33} implies that $$y_2/y_1$$ is nonconstant on $$I$$, $$y_2$$ isn't a constant multiple of $$y_1$$ on $$(a,b)$$. A similar argument shows that $$y_1$$ isn't a constant multiple of $$y_2$$ on $$(a,b)$$, since

\begin{eqnarray*}
\left(y_1\over y_2\right)'={y_1'y_2-y_1y_2'\over y_2^2}=-{W\over y_2^2}
\end{eqnarray*}

on any subinterval of $$(a,b)$$ where $$y_2$$ has no zeros.

We can now complete the proof of Theorem $$(2.1.3)$$. From Theorem $$(2.1.5)$$, two solutions $$y_1$$ and $$y_2$$ of \eqref{eq:2.1.32} are linearly independent on $$(a,b)$$ if and only if $$W$$ has no zeros on $$(a,b)$$. From Theorem $$(2.1.4)$$ and the motivating comments preceding it, $$\{y_1,y_2\}$$ is a fundamental set of solutions of \eqref{eq:2.1.32} if and only if $$W$$ has no zeros on $$(a,b)$$. Therefore $$\{y_1,y_2\}$$ is a fundamental set for \eqref{eq:2.1.32} on $$(a,b)$$ if and only if $$\{y_1,y_2\}$$ is linearly independent on $$(a,b)$$.

The next theorem summarizes the relationships among the concepts discussed in this section.

Theorem $$\PageIndex{6}$$

Suppose $$p$$ and $$q$$ are continuous on an open interval $$(a,b)$$ and let $$y_1$$ and $$y_2$$ be solutions of

\label{eq:2.1.34}
y''+p(x)y'+q(x)y=0

on $$(a,b).$$ Then the following statements are equivalent$$;$$ that is$$,$$ they are either all true or all false$$.$$

(a) The general solution of $$\eqref{eq:2.1.34}$$ on $$(a,b)$$ is $$y=c_1y_1+c_2y_2$$.

(b) $$\{y_1,y_2\}$$ is a fundamental set of solutions of $$\eqref{eq:2.1.34}$$ on $$(a,b).$$

(c) $$\{y_1,y_2\}$$ is linearly independent on $$(a,b).$$

(d) The Wronskian of $$\{y_1,y_2\}$$ is nonzero at some point in $$(a,b).$$

(e) The Wronskian of $$\{y_1,y_2\}$$ is nonzero at all points in $$(a,b).$$

Proof

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We can apply this theorem to an equation written as

\begin{eqnarray*}
P_0(x)y''+P_1(x)y'+P_2(x)y=0
\end{eqnarray*}

on an interval $$(a,b)$$ where $$P_0$$, $$P_1$$, and $$P_2$$ are continuous and $$P_0$$ has no zeros.

Theorem $$\PageIndex{7}$$

Suppose $$c$$ is in $$(a,b)$$ and $$\alpha$$ and $$\beta$$ are real numbers, not both zero. Under the assumptions of Theorem $$(2.1.7)$$, suppose $$y_{1}$$ and $$y_{2}$$ are solutions of \eqref{eq:2.1.34} such that

\label{eq:2.1.35}
\alpha y_{1}(c)+\beta y_{1}'(c)=0\text{\; and\; \; }
\alpha y_{2}(c)+\beta y_{2}'(c)=0.

Then $$\{y_{1},y_{2}\}$$ isn't linearly independent on $$(a,b).$$

Proof

Since $$\alpha$$ and $$\beta$$ are not both zero, \eqref{eq:2.1.35} implies that

\begin{eqnarray*}
\left|\begin{array}{ccccccc}
y_{1}(c)&y_{1}'(c)\\y_{2}(c)& y_{2}'(c)
\end{array}\right|=0, \text{\; so\; \; }
\left|\begin{array}{cccccc}
y_{1}(c)&y_{2}(c)\\ y_{1}'(c)&y_{2}'(c)
\end{array}\right|=0
\end{eqnarray*}

and Theorem $$(2.1.6)$$ implies the stated conclusion.
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