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2.1: Linear Second Order Homogeneous Equations

This page is a draft and is under active development. 

( \newcommand{\kernel}{\mathrm{null}\,}\)

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

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 (???) is homogeneous if f0 or nonhomogeneous if f. 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).

(c) Solve the initial value problem

\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*}

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

\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).

(c) Solve the initial value problem

\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*}

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

\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).

(c) Solve the initial value problem

\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*}

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

\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

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

\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).

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}

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

\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

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

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.

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

\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).

(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

\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.


This page titled 2.1: Linear Second Order Homogeneous Equations is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by William F. Trench.

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