2.1: Linear Second Order Homogeneous Equations

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$trench-1995.jpg$
Contributed by William F. Trench
Andrew G. Cowles Distinguished Professor Emeritus (Mathamatics) at Trinity University

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

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

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

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

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

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

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

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

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 $(-\infty,\infty)$.

\begin{equation}\label{eq:2.1.5}
y''-y=0,\quad y(0)=1,\quad y'(0)=3.
\end{equation}

Answer

(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

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

then

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

and

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

\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

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

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 $(-\infty,\infty)$.

\begin{equation}\label{eq:2.1.9}
y''+\omega^2y=0,\quad y(0)=1,\quad y'(0)=3.
\end{equation}

Answer

(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

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

then

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

and

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

\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

\begin{equation}\label{eq:2.1.12}
P_0(x)y''+P_1(x)y'+P_2(x)y=0,\quad y(x_0)=k_0,\quad y'(x_0)=k_1
\end{equation}

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

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

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*}
x^2y''+xy'-4y=0,\quad y(x_0)=k_0,\quad y'(x_0)=k_1
\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 $(0,\infty)$.

\begin{equation}\label{eq:2.1.14}
x^2y''+xy'-4y=0,\quad y(1)=2,\quad y'(1)=0.
\end{equation}

(d) Solve the initial value problem

\begin{equation}\label{eq:2.1.15}
x^2y''+xy'-4y=0,\quad y(-1)=2,\quad y'(-1)=0.
\end{equation}

Answer

(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

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

then

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

and

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

\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

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

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

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

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

Proof

\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

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

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

\begin{equation}\label{eq:2.1.21}
y''+p(x)y'+q(x)y=0,\quad y(x_0)=k_0,\quad y'(x_0)=k_1;
\end{equation}

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

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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.
\end{equation}

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

\begin{equation}\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.
\end{equation}

\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

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

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

\begin{equation}\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}
\end{equation}

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

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

on $(a,b)$, and define

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

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

\begin{equation} \label{eq:2.1.29}
W(x)=W(x_0) e^{-\int^x_{x_0}p(t)\, dt}, \quad a<x<b.
\end{equation}

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

Proof

Differentiating \eqref{eq:2.1.28} yields

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

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*}
y'+p(x)y=0,\quad y(x_0)=W(x_0).
\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) $y''+\omega^2y=0;\quad \quad y_1=\cos\omega x,\; y_2=\sin\omega x$

Answer

(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 $x$.

\begin{equation}\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}.
\end{equation}

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

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

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

\begin{equation}\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}.
\end{equation}

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*}
y''+p(x)y'+q(x)y=0,\quad y(x_0)=0,\quad y'(x_0)=0,
\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

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

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 $(a,b).$

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

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

\begin{equation} \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.
\end{equation}

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.