# 5.1: Homogeneous Linear Equations

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- William F. Trench
- Andrew G. Cowles Distinguished Professor Emeritus (Mathematics) at Trinity University

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

\[\label{eq:5.1.1} y''+p(x)y'+q(x)y=f(x).\]

We call the function \(f\) on the right a *forcing function*, since in physical applications it is often related to a force acting on some system modeled by the differential equation. We say that Equation \ref{eq:5.1.1} is *homogeneous* if \(f\equiv0\) or *nonhomogeneous* if \(f\not\equiv0\). Since these definitions are like the corresponding definitions in Section 2.1 for the linear first order equation

\[\label{eq:5.1.2} y'+p(x)y=f(x),\]

it is natural to expect similarities between methods of solving Equation \ref{eq:5.1.1} and Equation \ref{eq:5.1.2}. However, solving Equation \ref{eq:5.1.1} is more difficult than solving Equation \ref{eq:5.1.2}. For example, while Theorem 5.1.1 gives a formula for the general solution of Equation \ref{eq:5.1.2} in the case where \(f\equiv0\) and Theorem 5.1.2 gives a formula for the case where \(f\not\equiv0\), there are no formulas for the general solution of Equation \ref{eq:5.1.1} 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 is still necessary to solve the homogeneous equation

\[\label{eq:5.1.3} y''+p(x)y'+q(x)y=0\]

in order to solve the nonhomogeneous equation Equation \ref{eq:5.1.1}. This section is devoted to Equation \ref{eq:5.1.3}.

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

##### Theorem 5.1.1

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

\[y''+p(x)y'+q(x)y=0,\ y(x_0)=k_0,\ y'(x_0)=k_1 \nonumber \]

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

Since \(y\equiv0\) is obviously a solution of Equation \ref{eq:5.1.3} we call it the *trivial* solution. Any other solution is *nontrivial*. Under the assumptions of Theorem 5.1.1
, the only solution of the initial value problem

\[y''+p(x)y'+q(x)y=0,\ y(x_0)=0,\ y'(x_0)=0 \nonumber \]

on \((a,b)\) is the trivial solution (*Exercise 5.1.24*).

The next three examples illustrate concepts that we’ll develop later in this section. You shouldn’t be concerned with how to *find* the given solutions of the equations in these examples. This will be explained in later sections.

##### Example 5.1.1

The coefficients of \(y'\) and \(y\) in

\[\label{eq:5.1.4} y''-y=0\]

are the constant functions \(p\equiv0\) and \(q\equiv-1\), which are continuous on \((-\infty,\infty)\). Therefore Theorem 5.1.1 implies that every initial value problem for Equation \ref{eq:5.1.4} has a unique solution on \((-\infty,\infty)\).

- Verify that \(y_1=e^x\) and \(y_2=e^{-x}\) are solutions of Equation \ref{eq:5.1.4} on \((-\infty,\infty)\).
- Verify that if \(c_1\) and \(c_2\) are arbitrary constants, \(y=c_1e^x+c_2e^{-x}\) is a solution of Equation \ref{eq:5.1.4} on \((-\infty,\infty)\).
- Solve the initial value problem \[\label{eq:5.1.5} y''-y=0,\quad y(0)=1,\quad y'(0)=3.\]

**Solution**:

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 \[\label{eq:5.1.6} y=c_1e^x+c_2e^{-x}\] then \[\label{eq:5.1.7} y'=c_1e^x-c_2e^{-x}\] and \[y''=c_1e^x+c_2e^{-x},\nonumber \]

so \[\begin{aligned} y''-y&=(c_1e^x+c_2e^{-x})-(c_1e^x+c_2e^{-x})\\ &=c_1(e^x-e^x)+c_2(e^{-x}-e^{-x})=0\end{aligned}\nonumber \] for all \(x\). Therefore \(y=c_1e^x+c_2e^{-x}\) is a solution of Equation \ref{eq:5.1.4} on \((-\infty,\infty)\).

c.

We can solve Equation \ref{eq:5.1.5} by choosing \(c_1\) and \(c_2\) in Equation \ref{eq:5.1.6} so that \(y(0)=1\) and \(y'(0)=3\). Setting \(x=0\) in Equation \ref{eq:5.1.6} and Equation \ref{eq:5.1.7} shows that this is equivalent to

\[\begin{aligned} c_1+c_2&=1\\ c_1-c_2&=3.\end{aligned}\nonumber \]

Solving these equations yields \(c_1=2\) and \(c_2=-1\). Therefore \(y=2e^x-e^{-x}\) is the unique solution of Equation \ref{eq:5.1.5} on \((-\infty,\infty)\).

##### Example 5.1.2

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

\[\label{eq:5.1.8} y''+\omega^2y=0\]

are the constant functions \(p\equiv0\) and \(q\equiv\omega^2\), which are continuous on \((-\infty,\infty)\). Therefore Theorem 5.1.1 implies that every initial value problem for Equation \ref{eq:5.1.8} has a unique solution on \((-\infty,\infty)\).

- Verify that \(y_1=\cos\omega x\) and \(y_2=\sin\omega x\) are solutions of Equation \ref{eq:5.1.8} on \((-\infty,\infty)\).
- 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 Equation \ref{eq:5.1.8} on \((-\infty,\infty)\).
- Solve the initial value problem \[\label{eq:5.1.9} y''+\omega^2y=0,\quad y(0)=1,\quad y'(0)=3.\]

**Solution**:

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 \[\label{eq:5.1.10} y=c_1\cos\omega x+c_2\sin\omega x\] then \[\label{eq:5.1.11} y'=\omega(-c_1\sin\omega x+c_2\cos\omega x)\] and \[y''=-\omega^2(c_1\cos\omega x+c_2\sin\omega x),\nonumber \] so \[\begin{aligned} y''+\omega^2y&= -\omega^2(c_1\cos\omega x+c_2\sin\omega x) +\omega^2(c_1\cos\omega x+c_2\sin\omega x)\\ &=c_1\omega^2(-\cos\omega x+\cos\omega x)+ c_2\omega^2(-\sin\omega x+\sin\omega x)=0\end{aligned}\nonumber \] for all \(x\). Therefore \(y=c_1\cos\omega x+c_2\sin\omega x\) is a solution of Equation \ref{eq:5.1.8} on \((-\infty,\infty)\).

c. To solve Equation \ref{eq:5.1.9}, we must choosing \(c_1\) and \(c_2\) in Equation \ref{eq:5.1.10} so that \(y(0)=1\) and \(y'(0)=3\). Setting \(x=0\) in Equation \ref{eq:5.1.10} and Equation \ref{eq:5.1.11} shows that \(c_1=1\) and \(c_2=3/\omega\). Therefore

\[y=\cos\omega x+{3\over\omega}\sin\omega x\nonumber \]

is the unique solution of Equation \ref{eq:5.1.9} on \((-\infty,\infty)\).

Theorem 5.1.1 implies that if \(k_0\) and \(k_1\) are arbitrary real numbers then the initial value problem

\[\label{eq:5.1.12} P_0(x)y''+P_1(x)y'+P_2(x)y=0,\quad y(x_0)=k_0,\quad y'(x_0)=k_1\]

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 Equation \ref{eq:5.1.12} as

\[y''+{P_1(x)\over P_0(x)}y'+{P_2(x)\over P_0(x)}y=0\nonumber \]

and apply Theorem 5.1.1 with \(p=P_1/P_0\) and \(q=P_2/P_0\).

##### Example 5.1.3

The equation

\[\label{eq:5.1.13} x^2y''+xy'-4y=0\]

has the form of the differential equation in Equation \ref{eq:5.1.12}, 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 Equation \ref{eq:5.1.13} on \((-\infty,0)\) and \((0,\infty)\). Since \(P_0\) has no zeros on these intervals, Theorem 5.1.1 implies that the initial value problem

\[x^2y''+xy'-4y=0,\quad y(x_0)=k_0,\quad y'(x_0)=k_1\nonumber \]

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

- Verify that \(y_1=x^2\) is a solution of Equation \ref{eq:5.1.13} on \((-\infty,\infty)\) and \(y_2=1/x^2\) is a solution of Equation \ref{eq:5.1.13} on \((-\infty,0)\) and \((0,\infty)\).
- Verify that if \(c_1\) and \(c_2\) are any constants then \(y=c_1x^2+c_2/x^2\) is a solution of Equation \ref{eq:5.1.13} on \((-\infty,0)\) and \((0,\infty)\).
- Solve the initial value problem \[\label{eq:5.1.14} x^2y''+xy'-4y=0,\quad y(1)=2,\quad y'(1)=0.\]
- Solve the initial value problem \[\label{eq:5.1.15} x^2y''+xy'-4y=0,\quad y(-1)=2,\quad y'(-1)=0.\]

**Solution**:

a. If \(y_1=x^2\) then \(y_1'=2x\) and \(y_1''=2\), so \[x^2y_1''+xy_1'-4y_1=x^2(2)+x(2x)-4x^2=0\nonumber \] for \(x\) in \((-\infty,\infty)\). If \(y_2=1/x^2\), then \(y_2'=-2/x^3\) and \(y_2''=6/x^4\), so \[x^2y_2''+xy_2'-4y_2=x^2\left(6\over x^4\right)-x\left(2\over x^3\right)-{4\over x^2}=0\nonumber \] for \(x\) in \((-\infty,0)\) or \((0,\infty)\).

b. If \[\label{eq:5.1.16} y=c_1x^2+{c_2\over x^2}\] then \[\label{eq:5.1.17} y'=2c_1x-{2c_2\over x^3}\] and \[y''=2c_1+{6c_2\over x^4},\nonumber \] so \[\begin{aligned} x^{2}y''+xy'-4y&=x^{2}\left(2c_{1}+\frac{6c_{2}}{x^{4}} \right)+x\left(2c_{1}x-\frac{2c_{2}}{x^{3}} \right)-4\left(c_{1}x^{2}+\frac{c_{2}}{x^{2}} \right) \\ &=c_{1}(2x^{2}+2x^{2}-4x^{2})+c_{2}\left(\frac{6}{x^{2}}-\frac{2}{x^{2}}-\frac{4}{x^{2}} \right) \\ &=c_{1}\cdot 0+c_{2}\cdot 0 = 0 \end{aligned}\nonumber \] for \(x\) in \((-\infty,0)\) or \((0,\infty)\).

c. To solve Equation \ref{eq:5.1.14}, we choose \(c_1\) and \(c_2\) in Equation \ref{eq:5.1.16} so that \(y(1)=2\) and \(y'(1)=0\). Setting \(x=1\) in Equation \ref{eq:5.1.16} and Equation \ref{eq:5.1.17} shows that this is equivalent to

\[\begin{aligned} \phantom{2}c_1+\phantom{2}c_2&=2\\ 2c_1-2c_2&=0.\end{aligned}\nonumber \]

Solving these equations yields \(c_1=1\) and \(c_2=1\). Therefore \(y=x^2+1/x^2\) is the unique solution of Equation \ref{eq:5.1.14} on \((0,\infty)\).

d. We can solve Equation \ref{eq:5.1.15} by choosing \(c_1\) and \(c_2\) in Equation \ref{eq:5.1.16} so that \(y(-1)=2\) and \(y'(-1)=0\). Setting \(x=-1\) in Equation \ref{eq:5.1.16} and Equation \ref{eq:5.1.17} shows that this is equivalent to

\[\begin{aligned} \phantom{-2}c_1+\phantom{2}c_2&=2\\ -2c_1+2c_2&=0.\end{aligned}\nonumber \]

Solving these equations yields \(c_1=1\) and \(c_2=1\). Therefore \(y=x^2+1/x^2\) is the unique solution of Equation \ref{eq:5.1.15} on \((-\infty,0)\).

Although the *formulas* for the solutions of Equation \ref{eq:5.1.14} and Equation \ref{eq:5.1.15} 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 *on an interval that contains the initial point*; therefore, the solution of Equation \ref{eq:5.1.14} is \(y=x^2+1/x^2\) *on the interval* \((0,\infty)\), which contains the initial point \(x_0=1\), while the solution of Equation \ref{eq:5.1.15} is \(y=x^2+1/x^2\) *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

\[y=c_1y_1+c_2y_2\nonumber \]

is a *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 5.1.1 , 5.1.2 , 5.1.3 .

##### Theorem 5.1.2

If \(y_1\) and \(y_2\) are solutions of the homogeneous equation

\[\label{eq:5.1.18} y''+p(x)y'+q(x)y=0\]

on \((a,b),\) then any linear combination\[\label{eq:5.1.19} y=c_1y_1+c_2y_2\]

of \(y_1\) and \(y_2\) is also a solution of \(\eqref{eq:5.1.18}\) on \((a,b).\)

**Proof**-
If \[y=c_1y_1+c_2y_2\nonumber \] then \[y'=c_1y_1'+c_2y_2'\quad\text{ and} \quad y''=c_1y_1''+c_2y_2''.\nonumber \]

Therefore\[\begin{aligned} y''+p(x)y'+q(x)y&=(c_1y_1''+c_2y_2'')+p(x)(c_1y_1'+c_2y_2') +q(x)(c_1y_1+c_2y_2)\\ &=c_1\left(y_1''+p(x)y_1'+q(x)y_1\right) +c_2\left(y_2''+p(x)y_2'+q(x)y_2\right)\\ &=c_1\cdot0+c_2\cdot0=0,\end{aligned}\nonumber \]

since \(y_1\) and \(y_2\) are solutions of Equation \ref{eq:5.1.18}.We say that \(\{y_1,y_2\}\) is a

*fundamental set of solutions of \(\eqref{eq:5.1.18}\)*on \((a,b)\) if every solution of Equation \ref{eq:5.1.18} on \((a,b)\) can be written as a linear combination of \(y_1\) and \(y_2\) as in Equation \ref{eq:5.1.19}. In this case we say that Equation \ref{eq:5.1.19} is*general solution of \(\eqref{eq:5.1.18}\)*on \((a,b)\).

## Linear Independence

We need a way to determine whether a given set \(\{y_1,y_2\}\) of solutions of Equation \ref{eq:5.1.18} 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 *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 Equation \ref{eq:5.1.18}, since, for example, if \(y_1\equiv0\) we could write \(y_1=0y_2\).) We’ll also say that the set \(\{y_1,y_2\}\) *is linearly independent on* \((a,b)\).

##### Theorem 5.1.3

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

\[\label{eq:5.1.20} y''+p(x)y'+q(x)y=0\]

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

**Proof**-
We’ll present the proof of Theorem 5.1.3 in steps worth regarding as theorems in their own right. However, let’s first interpret Theorem 5.1.3 in terms of Examples 5.1.1 , 5.1.2 , 5.1.3 .

##### Example 5.1.4

Since \(e^x/e^{-x}=e^{2x}\) is nonconstant, Theorem 5.1.3 implies that \(y=c_1e^x+c_2e^{-x}\) is the general solution of \(y''-y=0\) on \((-\infty,\infty)\).

Since \(\cos\omega x/\sin\omega x=\cot\omega x\) is nonconstant, Theorem 5.1.3 implies that \(y=c_1\cos\omega x+c_2\sin\omega x\) is the general solution of \(y''+\omega^2y=0\) on \((-\infty,\infty)\).

Since \(x^2/x^{-2}=x^4\) is nonconstant, Theorem 5.1.3 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 5.1.3 , let’s see what is required to prove that \(\{y_1,y_2\}\) is a fundamental set of solutions of Equation \ref{eq:5.1.20} on \((a,b)\). Let \(x_0\) be an arbitrary point in \((a,b)\), and suppose \(y\) is an arbitrary solution of Equation \ref{eq:5.1.20} on \((a,b)\). Then \(y\) is the unique solution of the initial value problem

\[\label{eq:5.1.21} y''+p(x)y'+q(x)y=0,\quad y(x_0)=k_0,\quad y'(x_0)=k_1;\]

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 5.1.1 implies that Equation \ref{eq:5.1.21} 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 Equation \ref{eq:5.1.20} on \((a,b)\) if and only if it is possible to write the solution of an arbitrary initial value problem Equation \ref{eq:5.1.21} as \(y=c_1y_1+c_2y_2\). This is equivalent to requiring that the system\[\label{eq:5.1.22} \begin{array}{rcl} c_1y_1(x_0)+c_2y_2(x_0)&=k_0\\ c_1y_1'(x_0)+c_2y_2'(x_0)&=k_1 \end{array}\]

has a solution \((c_1,c_2)\) for every choice of \((k_0,k_1)\). Let’s try to solve Equation \ref{eq:5.1.22}.Multiplying the first equation in Equation \ref{eq:5.1.22} by \(y_2'(x_0)\) and the second by \(y_2(x_0)\) yields

\[\begin{aligned} c_1y_1(x_0)y_2'(x_0)+c_2y_2(x_0)y_2'(x_0)&= y_2'(x_0)k_0\\ c_1y_1'(x_0)y_2(x_0)+c_2y_2'(x_0)y_2(x_0)&= y_2(x_0)k_1,\end{aligned}\]

and subtracting the second equation here from the first yields\[\label{eq:5.1.23} \left(y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0)\right)c_1= y_2'(x_0)k_0-y_2(x_0)k_1.\]

Multiplying the first equation in Equation \ref{eq:5.1.22} by \(y_1'(x_0)\) and the second by \(y_1(x_0)\) yields\[\begin{aligned} c_1y_1(x_0)y_1'(x_0)+c_2y_2(x_0)y_1'(x_0)&= y_1'(x_0)k_0\\ c_1y_1'(x_0)y_1(x_0)+c_2y_2'(x_0)y_1(x_0)&= y_1(x_0)k_1,\end{aligned}\]

and subtracting the first equation here from the second yields\[\label{eq:5.1.24} \left(y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0)\right)c_2= y_1(x_0)k_1-y_1'(x_0)k_0.\]

If\[y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0)=0,\nonumber \]

it is impossible to satisfy Equation \ref{eq:5.1.23} and Equation \ref{eq:5.1.24} (and therefore Equation \ref{eq:5.1.22} ) unless \(k_0\) and \(k_1\) happen to satisfy\[\begin{aligned} y_1(x_0)k_1-y_1'(x_0)k_0&=0\\ y_2'(x_0)k_0-y_2(x_0)k_1&=0.\end{aligned}\]

On the other hand, if\[\label{eq:5.1.25} y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0)\ne0\]

we can divide Equation \ref{eq:5.1.23} and Equation \ref{eq:5.1.24} through by the quantity on the left to obtain\[\label{eq:5.1.26} \begin{array}{rcl} c_1&={y_2'(x_0)k_0-y_2(x_0)k_1\over y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0)}\\ c_2&={y_1(x_0)k_1-y_1'(x_0)k_0\over y_1(x_0)y_2'(x_0)-y_1'(x_0)y_2(x_0)}, \end{array}\]

no matter how \(k_0\) and \(k_1\) are chosen. This motivates us to consider conditions on \(y_1\) and \(y_2\) that imply Equation \ref{eq:5.1.25}.

##### Theorem 5.1.4

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

\[\label{eq:5.1.27} y''+p(x)y'+q(x)y=0\]

on \((a,b)\), and define\[\label{eq:5.1.28} W=y_1y_2'-y_1'y_2.\]

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

\[\label{eq:5.1.29} W(x)=W(x_0) e^{-\int^x_{x_0}p(t)\:dt}, \quad a<x<b\]

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

**Proof**-
Differentiating Equation \ref{eq:5.1.28} yields

\[\label{eq:5.1.30} W'=y'_1y'_2+y_1y''_2-y'_1y'_2-y''_1y_2= y_1y''_2-y''_1y_2.\]

Since \(y_1\) and \(y_2\) both satisfy Equation \ref{eq:5.1.27},\[y''_1 =-py'_1-qy_1\quad \text{and} \quad y''_2 =-py'_2-qy_2.\nonumber \]

Substituting these into Equation \ref{eq:5.1.30} yields\[\begin{aligned} W'&= -y_1\bigl(py'_2+qy_2\bigr) +y_2\bigl(py'_1+qy_1\bigr) \\ &= -p(y_1y'_2-y_2y'_1)-q(y_1y_2-y_2y_1)\\ &= -p(y_1y'_2-y_2y'_1)=-pW.\end{aligned}\nonumber \]

Therefore \(W'+p(x)W=0\); that is, \(W\) is the solution of the initial value problem\[y'+p(x)y=0,\quad y(x_0)=W(x_0).\nonumber \]

We leave it to you to verify by separation of variables that this implies Equation \ref{eq:5.1.29}. If \(W(x_0)\ne0\), Equation \ref{eq:5.1.29} implies that \(W\) has no zeros in \((a,b)\), since an exponential is never zero. On the other hand, if \(W(x_0)=0\), Equation \ref{eq:5.1.29} implies that \(W(x)=0\) for all \(x\) in \((a,b)\).

The function \(W\) defined in Equation \ref{eq:5.1.28} is the *Wronskian of \(\{y_1,y_2\}\)*. Formula Equation \ref{eq:5.1.29} is *Abel’s formula*.

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

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

The expressions in Equation \ref{eq:5.1.26} for \(c_1\) and \(c_2\) can be written in terms of determinants as\[c_1={1\over W(x_0)} \left| \begin{array}{cc} k_0 & y_2(x_0) \\ k_1 & y'_2(x_0) \end{array} \right| \quad \text{and} \quad c_2={1\over W(x_0)} \left| \begin{array}{cc} y_1(x_0) & k_0 \\ y'_1(x_0) &k_1 \end{array} \right|.\nonumber \]

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

##### Example 5.1.5

Verify Abel’s formula for the following differential equations and the corresponding solutions, from Examples 5.1.1 , 5.1.2 , 5.1.3 .

- \(y''-y=0;\quad y_1=e^x,\; y_2=e^{-x}\)
- \(y''+\omega^2y=0;\quad y_1=\cos\omega x,\; y_2=\sin\omega x\)
- \(x^2y''+xy'-4y=0;\quad y_1=x^2,\; y_2=1/x^2\)

**Solution**:

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

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

for all \(x\).

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

\[\begin{aligned} W(x)&={\left| \begin{array}{cc} \cos\omega x & \sin\omega x \\ -\omega\sin\omega x &\omega\cos\omega x \end{array} \right|}\\ &=\cos\omega x (\omega\cos\omega x)-(-\omega\sin\omega x)\sin\omega x\\ &=\omega(\cos^2\omega x+\sin^2\omega x)=\omega\end{aligned}\nonumber \]

for all \(x\).

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

\[\label{eq:5.1.31} W=\left| \begin{array}{cc} x^2 & 1/x^2 \\ 2x & -2/x^3 \end{array} \right|=x^2\left(-{2\over x^3}\right)-2x\left(1\over x^2\right)=-{4\over x}.\]

To verify Abel’s formula we rewrite the differential equation as\[y''+{1\over x}y'-{4\over x^2}y=0\nonumber \]

to see that \(p(x)=1/x\). If \(x_0\) and \(x\) are either both in \((-\infty,0)\) or both in \((0,\infty)\) then\[\int_{x_0}^x p(t)\,dt=\int_{x_0}^x {dt\over t}=\ln\left(x\over x_0\right),\nonumber \]

so Abel’s formula becomes\[\begin{aligned} W(x)&=W(x_0)e^{-\ln(x/x_0)}=W(x_0){x_0\over x}\\ &=-\left(4\over x_0\right)\left(x_0\over x\right)\quad \text{from} \eqref{eq:5.1.31}\\ &=-{4\over x},\end{aligned}\nonumber \]

which is consistent with Equation \ref{eq:5.1.31}.

The next theorem will enable us to complete the proof of Theorem 5.1.3 .

##### Theorem 5.1.5

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

\[\label{eq:5.1.32} y''+p(x)y'+q(x)y=0\]

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 are finished with this part of the proof.) Then \(y_2/y_1\) is defined on \(I\), and

\[\label{eq:5.1.33} \left(y_2\over y_1\right)'={y_1y_2'-y_1'y_2\over y_1^2}={W\over y_1^2}.\]

However, if \(W(x_0)=0\), Theorem 5.1.4 implies that \(W\equiv0\) on \((a,b)\). Therefore Equation \ref{eq:5.1.33} 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 Equation \ref{eq:5.1.32} 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

\[y''+p(x)y'+q(x)y=0,\quad y(x_0)=0,\quad y'(x_0)=0,\nonumber \]

which implies that \(Y\equiv0\) on \((a,b)\), by the paragraph following Theorem 5.1.1 . (See also

*Exercise 5.1.24*). 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 Equation \ref{eq:5.1.33} 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

\[\left(y_1\over y_2\right)'={y_1'y_2-y_1y_2'\over y_2^2}=-{W\over y_2^2}\nonumber \]

on any subinterval of \((a,b)\) where \(y_2\) has no zeros.We can now complete the proof of Theorem 5.1.3 . From Theorem 5.1.5 , two solutions \(y_1\) and \(y_2\) of Equation \ref{eq:5.1.32} are linearly independent on \((a,b)\) if and only if \(W\) has no zeros on \((a,b)\). From Theorem 5.1.4 and the motivating comments preceding it, \(\{y_1,y_2\}\) is a fundamental set of solutions of Equation \ref{eq:5.1.32} if and only if \(W\) has no zeros on \((a,b)\). Therefore \(\{y_1,y_2\}\) is a fundamental set for Equation \ref{eq:5.1.32} 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 5.1.6

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

\[\label{eq:5.1.34} y''+p(x)y'+q(x)y=0\]

on \((a,b).\) Then the following statements are equivalent\(;\) that is\(,\) they are either all true or all false\(.\)- The general solution of \(\eqref{eq:5.1.34}\) on \((a,b)\) is \(y=c_1y_1+c_2y_2\).
- \(\{y_1,y_2\}\) is a fundamental set of solutions of \(\eqref{eq:5.1.34}\) on \((a,b).\)
- \(\{y_1,y_2\}\) is linearly independent on \((a,b).\)
- The Wronskian of \(\{y_1,y_2\}\) is nonzero at some point in \((a,b).\)
- The Wronskian of \(\{y_1,y_2\}\) is nonzero at all points in \((a,b).\)

We can apply this theorem to an equation written as

\[P_0(x)y''+P_1(x)y'+P_2(x)y=0\nonumber \]

on an interval \((a,b)\) where \(P_0\), \(P_1\), and \(P_2\) are continuous and \(P_0\) has no zeros.dd proof here and it will automatically be hidden

##### Theorem 5.1.7

Suppose \(c\) is in \((a,b)\) and \(\alpha\) and \(\beta\) are real numbers, not both zero. Under the assumptions of Theorem 5.1.7 , suppose \(y_{1}\) and \(y_{2}\) are solutions of Equation \ref{eq:5.1.34} such that

\[\label{eq:5.1.35} \alpha y_{1}(c)+\beta y_{1}'(c)=0\quad\text{and}\quad \alpha y_{2}(c)+\beta y_{2}'(c)=0.\]

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

**Proof**-
Since \(\alpha\) and \(\beta\) are not both zero, Equation \ref{eq:5.1.35} implies that

\[\left|\begin{array}{ccccccc} y_{1}(c)&y_{1}'(c)\\y_{2}(c)& y_{2}'(c) \end{array}\right|=0, \quad\text{so}\quad \left|\begin{array}{cccccc} y_{1}(c)&y_{2}(c)\\ y_{1}'(c)&y_{2}'(c) \end{array}\right|=0\nonumber \]

and Theorem 5.1.6 implies the stated conclusion.