5.1 Homogeneous Linear Equations

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

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 (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

it’s natural to expect similarities between methods of solving (5.1.1) and (5.1.2). However, solving (5.1.1) is more difficult than solving (5.1.2). For example, while Theorem 2.1.1 gives a formula for the general solution of (5.1.2) in the case where \(f\equiv0\) and Theorem 2.1.2 gives a formula for the case where \(f\not\equiv0\), there are no formulas for the general solution of (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’s still necessary to solve the homogeneous equation

in order to solve the nonhomogeneous equation (5.1.1). This section is devoted to (5.1.3).

The next theorem gives sufficient conditions for existence and uniqueness of solutions of initial value problems for (5.1.3). We omit the proof.

Since \(y\equiv0\) is obviously a solution of (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

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

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 \(y_2''-y_2=0\).

Solution (b) If

then

and

so

for all \(x\). Therefore \(y=c_1e^x+c_2e^{-x}\) is a solution of (5.1.4) on \((-\infty,\infty)\).

Solution (c) We can solve (5.1.5) by choosing \(c_1\) and \(c_2\) in (5.1.6) so that \(y(0)=1\) and \(y'(0)=3\). Setting \(x=0\) in (5.1.6) and (5.1.7) shows that this is equivalent to

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

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 \(y_2''+\omega^2y_2=0\).

Solution (b) If

then

and

so

for all \(x\). Therefore \(y=c_1\cos\omega x+c_2\sin\omega x\) is a solution of (5.1.8) on \((-\infty,\infty)\).

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

is the unique solution of (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

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 (5.1.12) as

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

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

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

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

Solution (b) If

then

and

so

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

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

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

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

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

Although the formulas for the solutions of (5.1.14) and (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 (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 (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

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, and 5.1.3.

Proof If

then

Therefore

since \(y_1\) and \(y_2\) are solutions of (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 (5.1.18) on \((a,b)\) can be written as a linear combination of \(y_1\) and \(y_2\) as in (5.1.19). In this case we say that (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 (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 (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 \((a,b)\).

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, and 5.1.3.

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 (5.1.20) on \((a,b)\). Let \(x_0\) be an arbitrary point in \((a,b)\), and suppose \(y\) is an arbitrary solution of (5.1.20) on \((a,b)\). Then \(y\) is the unique solution of the initial value problem

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 (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 (5.1.20) on \((a,b)\) if and only if it’s possible to write the solution of an arbitrary initial value problem (5.1.21) as \(y=c_1y_1+c_2y_2\). This is equivalent to requiring that the system

has a solution \((c_1,c_2)\) for every choice of \((k_0,k_1)\). Let’s try to solve (5.1.22).

Multiplying the first equation in (5.1.22) by \(y_2'(x_0)\) and the second by \(y_2(x_0)\) yields

and subtracting the second equation here from the first yields

Multiplying the first equation in (5.1.22) by \(y_1'(x_0)\) and the second by \(y_1(x_0)\) yields

and subtracting the first equation here from the second yields

If

it’s impossible to satisfy (5.1.23) and (5.1.24) (and therefore (5.1.22)) unless \(k_0\) and \(k_1\) happen to satisfy

On the other hand, if

we can divide (5.1.23) and (5.1.24) through by the quantity on the left to obtain

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

Proof Differentiating (5.1.28) yields

Since \(y_1\) and \(y_2\) both satisfy (5.1.27),

Substituting these into (5.1.30) yields

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

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

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

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

The expressions in (5.1.26) for \(c_1\) and \(c_2\) can be written in terms of determinants as

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

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

for all \(x\).

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

for all \(x\).

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

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

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

so Abel’s formula becomes

which is consistent with (5.1.31).∎

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

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

However, if \(W(x_0)=0\), Theorem 5.1.4 implies that \(W\equiv0\) on \((a,b)\). Therefore (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 (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

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

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 (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 (5.1.32) if and only if \(W\) has no zeros on \((a,b)\). Therefore \(\{y_1,y_2\}\) is a fundamental set for (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.

We can apply this theorem to an equation written as

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

Proof Since \(\alpha\) and \(\beta\) are not both zero, (5.1.35) implies that

and Theorem 5.1.6 implies the stated conclusion.

5.1 Exercises

1. 1. Verify that \(y_1=e^{2x}\) and \(y_2=e^{5x}\) are solutions of

      \[ y''-7y'+10y=0 \tag*{\rm (A)} \]

      on \((-\infty,\infty)\).

   2. Verify that if \(c_1\) and \(c_2\) are arbitrary constants then \(y=c_1e^{2x}+c_2e^{5x}\) is a solution of (A) on \((-\infty,\infty)\).

   3. Solve the initial value problem

      \[ y''-7y'+10y=0,\quad y(0)=-1,\quad y'(0)=1. \]

   4. Solve the initial value problem

      \[ y''-7y'+10y=0,\quad y(0)=k_0,\quad y'(0)=k_1. \]
   Show answer

   (c) \(y=-2e^{2x}+e^{5x}\) (d) \(y=(5k_0-k_1)\dst{e^{2x}\over3}+(k_1-2k_0)\dst{e^{5x}\over3}\).

2. 1. Verify that \(y_1=e^x\cos x\) and \(y_2=e^x\sin x\) are solutions of

      \[ y''-2y'+2y=0 \tag*{\rm (A)} \]

      on \((-\infty,\infty)\).

   2. Verify that if \(c_1\) and \(c_2\) are arbitrary constants then \(y=c_1e^x\cos x+c_2e^x\sin x\) is a solution of (A) on \((-\infty,\infty)\).

   3. Solve the initial value problem

      \[ y''-2y'+2y=0,\quad y(0)=3,\quad y'(0)=-2. \]

   4. Solve the initial value problem

      \[ y''-2y'+2y=0,\quad y(0)=k_0,\quad y'(0)=k_1. \]
   Show answer

   (c) \(y=e^x(3\cos x-5\sin x)\) (d) \(y=e^x\left(k_0\cos x+(k_1-k_0)\sin x\right)\)

3. 1. Verify that \(y_1=e^x\) and \(y_2=xe^x\) are solutions of

      \[ y''-2y'+y=0 \tag*{\rm (A)} \]

      on \((-\infty,\infty)\).

   2. Verify that if \(c_1\) and \(c_2\) are arbitrary constants then \(y=e^x(c_1+c_2x)\) is a solution of (A) on \((-\infty,\infty)\).

   3. Solve the initial value problem

      \[ y''-2y'+y=0,\quad y(0)=7,\quad y'(0)=4. \]

   4. Solve the initial value problem

      \[ y''-2y'+y=0,\quad y(0)=k_0,\quad y'(0)=k_1. \]
   Show answer

   (c) \(y=e^x(7-3x)\) (d) \(y=e^x\left(k_0+(k_1-k_0)x\right)\)

4. 1. Verify that \(y_1=1/(x-1)\) and \(y_2=1/(x+1)\) are solutions of

      \[ (x^2-1)y''+4xy'+2y=0 \tag*{\rm (A)} \]

      on \((-\infty,-1)\), \((-1,1)\), and \((1,\infty)\). What is the general solution of (A) on each of these intervals?

   2. Solve the initial value problem

      \[ (x^2-1)y''+4xy'+2y=0,\quad y(0)=-5,\quad y'(0)=1. \]

      What is the interval of validity of the solution?

   3. C/G Graph the solution of the initial value problem.

   4. Verify Abel’s formula for \(y_1\) and \(y_2\), with \(x_0=0\).

   Show answer

   (a) \(y=\dst{c_1\over x-1}+\dst{c_2\over x+1}\) (b) \(y=\dst{{2\over x-1}-{3\over x+1}};\; (-1,1)\)

5. Compute the Wronskians of the given sets of functions.

   (a) \(\{1, e^x\}\)

   (b) \(\{e^x, e^x \sin x\}\)

   (c) \(\{x+1, x^2+2\}\)

   (d) \(\{ x^{1/2}, x^{-1/3}\}\)

   (e) \(\{\dst \frac{\sin x}{x}, \frac{\cos x}{x}\}\)	(f) \(\{ x \ln|x|, x^2\ln|x|\}\)
   (g) \(\{e^x\cos\sqrt x, e^x\sin\sqrt x\}\)

   Show answer

   (a) \(e^x\) (b) \(e^{2x}\cos x\) (c) \(x^2+2x-2\) (d) \(\displaystyle{-{5\over6}x^{-5/6}}\) (e) \(\displaystyle{-\frac{1}{x^2}}\) (f) \((x\ln|x|)^2\) (g) \(\dst{e^{2x}\over2\sqrt x}\)

6. Find the Wronskian of a given set \(\{y_1,y_2\}\) of solutions of

   \[ y''+3(x^2+1)y'-2y=0, \]

   given that \(W(\pi)=0\).

   Show answer

   \(0\)

7. Find the Wronskian of a given set \(\{y_1,y_2\}\) of solutions of

   \[ (1-x^2)y''-2xy'+\alpha(\alpha+1)y=0, \]

   given that \(W(0)=1\). (This is Legendre’s equation.)

   Show answer

   \(W(x)=(1-x^2)^{-1}\)

8. Find the Wronskian of a given set \(\{y_1,y_2\}\) of solutions of

   \[ x^2y''+xy'+(x^2-\nu^2)y=0 , \]

   given that \(W(1)=1\). (This is Bessel’s equation.)

   Show answer

   \(\displaystyle{W(x)={1\over x}}\)

9. (This exercise shows that if you know one nontrivial solution of \(y''+p(x)y'+q(x)y=0\), you can use Abel’s formula to find another.)

   Suppose \(p\) and \(q\) are continuous and \(y_1\) is a solution of

   \[ y''+p(x)y'+q(x)y=0 \tag*{\rm (A)} \]

   that has no zeros on \((a,b)\). Let \(P(x)=\int p(x)\,dx\) be any antiderivative of \(p\) on \((a,b)\).

   1. Show that if \(K\) is an arbitrary nonzero constant and \(y_2\) satisfies

      \[ y_1y_2'-y_1'y_2=Ke^{-P(x)} \tag*{\rm (B)} \]

      on \((a,b)\), then \(y_2\) also satisfies (A) on \((a,b)\), and \(\{y_1,y_2\}\) is a fundamental set of solutions on (A) on \((a,b)\).

   2. Conclude from (a) that if \(y_2=uy_1\) where \(u'=K\dst{e^{-P(x)}\over y_1^2(x)}\), then \(\{y_1,y_2\}\) is a fundamental set of solutions of (A) on \((a,b)\).

10. \(y''-2y'-3y=0\);   \(y_1=e^{3x}\)

    Show answer

    \(y_2=e^{-x}\)

11. \(y''-6y'+9y=0\);   \(y_1=e^{3x}\)

    Show answer

    \(y_2=xe^{3x}\)

12. \(y''-2ay'+a^2y=0\)  (\(a=\) constant);   \(y_1=e^{ax}\)

    Show answer

    \(y_2=xe^{ax}\)

13. \(x^2y''+xy'-y=0\);   \(y_1=x\)

    Show answer

    \(y_2=\dst{1\over x}\)

14. \(x^2y''-xy'+y=0\);   \(y_1=x\)

    Show answer

    \(y_2=x\ln x\)

15. \(x^2y''-(2a-1)xy'+a^2y=0\)  (\(a=\) nonzero constant);  \(x>0\);   \(y_1=x^a\)

    Show answer

    \(y_2=x^a\ln x\)

16. \(4x^2y''-4xy'+(3-16x^2)y=0\);   \(y_1=x^{1/2}e^{2x}\)

    Show answer

    \(y_2=x^{1/2}e^{-2x}\)

17. \((x-1)y''-xy'+y=0\);   \(y_1=e^x\)

    Show answer

    \(y_2=x\)

18. \(x^2y''-2xy'+(x^2+2)y=0\);   \(y_1=x\cos x\)

    Show answer

    \(y_2=x\sin x\)

19. \(4x^2(\sin x)y''-4x(x\cos x+\sin x)y'+(2x\cos x+3\sin x)y=0\);   \(y_1=x^{1/2}\)

    Show answer

    \(y_2=x^{1/2}\cos x\)

20. \((3x-1)y''-(3x+2)y'-(6x-8)y=0\);   \(y_1=e^{2x}\)

    Show answer

    \(y_2=xe^{-x}\)

21. \((x^2-4)y''+4xy'+2y=0\);   \(y_1=\dst{1\over x-2}\)

    Show answer

    \(y_2=\dst{1\over x^2-4}\)

22. \((2x+1)xy''-2(2x^2-1)y'-4(x+1)y=0\);  \(y_1=\dst{1\over x}\)

    Show answer

    \(y_2=e^{2x}\)

23. \((x^2-2x)y''+(2-x^2)y'+(2x-2)y=0\);  \(y_1=e^x\)

    Show answer

    \(y_2=x^2\)

24. Suppose \(p\) and \(q\) are continuous on an open interval \((a,b)\) and let \(x_0\) be in \((a,b)\). Use Theorem 5.1.1 to show that the only solution of the initial value problem

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

    on \((a,b)\) is the trivial solution \(y\equiv0\).

25. Suppose \(P_0\), \(P_1\), and \(P_2\) are continuous on \((a,b)\) and let \(x_0\) be in \((a,b)\). Show that if either of the following statements is true then \(P_0(x)=0\) for some \(x\) in \((a,b)\).

    1. The initial value problem

       \[ P_0(x)y''+P_1(x)y'+P_2(x)y=0,\quad y(x_0)=k_0,\quad y'(x_0)=k_1 \]

       has more than one solution on \((a,b)\).

    2. The initial value problem

       \[ P_0(x)y''+P_1(x)y'+P_2(x)y=0,\quad y(x_0)=0,\quad y'(x_0)=0 \]

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

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

    \[ y''+p(x)y'+q(x)y=0 \tag*{\rm (A)} \]

    on \((a,b)\). Let

    \[ z_1=\alpha y_1+\beta y_2\mbox{\quad and \quad} z_2=\gamma y_1+\delta y_2, \]

    where \(\alpha\), \(\beta\), \(\gamma\), and \(\delta\) are constants. Show that if \(\{z_1,z_2\}\) is a fundamental set of solutions of (A) on \((a,b)\) then so is \(\{y_1,y_2\}\).

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

    \[ y''+p(x)y'+q(x)y=0 \tag*{\rm (A)} \]

    on \((a,b)\). Let

    \[ z_1=\alpha y_1+\beta y_2\mbox{\quad and \quad} z_2=\gamma y_1+\delta y_2, \]

    where \(\alpha,\beta,\gamma\), and \(\delta\) are constants. Show that \(\{z_1,z_2\}\) is a fundamental set of solutions of (A) on \((a,b)\) if and only if \( \alpha\gamma-\beta\delta\ne0\).

28. Suppose \(y_1\) is differentiable on an interval \((a,b)\) and \(y_2=ky_1\), where \(k\) is a constant. Show that the Wronskian of \(\{y_1,y_2\}\) is identically zero on \((a,b)\).

29. Let

    \[ y_1=x^3\quad\mbox{ and }\quad y_2=\left\{\begin{array}{rl} x^3,&x\ge 0,\\ -x^3,&x<0.\end{array}\right. \]

    1. Show that the Wronskian of \(\{y_1,y_2\}\) is defined and identically zero on \((-\infty,\infty)\).

    2. Suppose \(a<0<b\). Show that \(\{y_1,y_2\}\) is linearly independent on \((a,b)\).

    3. Use Exercise 25(b) to show that these results don’t contradict Theorem 5.1.5, because neither \(y_1\) nor \(y_2\) can be a solution of an equation

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

       on \((a,b)\) if \(p\) and \(q\) are continuous on \((a,b)\).

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

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

    on \((a,b)\) such that either \(y_1(x_0)=y_2(x_0)=0\) or \(y_1'(x_0)=y_2'(x_0)=0\) for some \(x_0\) in \((a,b)\). Show that \(\{y_1,y_2\}\) is linearly dependent on \((a,b)\).

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

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

    on \((a,b)\). Show that if \(y_1(x_1)=y_1(x_2)=0\), where \(a<x_1<x_2<b\), then \(y_2(x)=0\) for some \(x\) in \((x_1,x_2)\). Hint: Show that if \(y_2\) has no zeros in \((x_1,x_2)\), then \(y_1/y_2\) is either strictly increasing or strictly decreasing on \((x_1,x_2)\), and deduce a contradiction.

32. Suppose \(p\) and \(q\) are continuous on \((a,b)\) and every solution of

    \[ y''+p(x)y'+q(x)y=0 \tag*{\rm (A)} \]

    on \((a,b)\) can be written as a linear combination of the twice differentiable functions \(\{y_1,y_2\}\). Use Theorem 5.1.1 to show that \(y_1\) and \(y_2\) are themselves solutions of (A) on \((a,b)\).

33. Suppose \(p_1\), \(p_2\), \(q_1\), and \(q_2\) are continuous on \((a,b)\) and the equations

    \[ y''+p_1(x)y'+q_1(x)y=0 \mbox{\quad and \quad} y''+p_2(x)y'+q_2(x)y=0 \]

    have the same solutions on \((a,b)\). Show that \(p_1=p_2\) and \(q_1=q_2\) on \((a,b)\). Hint: Use Abel’s formula.

34. (For this exercise you have to know about \(3\times 3\) determinants.) Show that if \(y_1\) and \(y_2\) are twice continuously differentiable on \((a,b)\) and the Wronskian \(W\) of \(\{y_1,y_2\}\) has no zeros in \((a,b)\) then the equation

    \[ \frac{1}{W} \left| \begin{array}{ccc} y & y_1 & y_2 \\[6pt] y' & y'_1 & y'_2 \\[6pt] y'' & y_1'' & y_2'' \end{array} \right|=0 \]

    can be written as

    \[ y''+p(x)y'+q(x)y=0, \tag*{\rm (A)} \]

    where \(p\) and \(q\) are continuous on \((a,b)\) and \(\{y_1,y_2\}\) is a fundamental set of solutions of (A) on \((a,b)\). Hint: Expand the determinant by cofactors of its first column.

35. Use the method suggested by Exercise 34 to find a linear homogeneous equation for which the given functions form a fundamental set of solutions on some interval.

    (a) \(e^x \cos 2x, \quad e^x \sin 2x\)

    (b) \(x, \quad e^{2x}\)

    (c) \(x, \quad x \ln x\)

    (d) \(\cos (\ln x), \quad \sin (\ln x)\)

    (e) \(\cosh x, \quad \sinh x\)

    (f) \( x^2-1, \quad x^2+1\)

    Show answer

    (a) \(y''-2y'+5y=0\) (b) \((2x-1)y''-4xy'+4y=0\) (c) \(x^2y''-xy'+y=0\)

         (d) \(x^2y''+xy'+y=0\) (e) \(y''-y=0\) (f) \(xy''-y'=0\)

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

    \[ y''+p(x)y'+q(x)y=0 \tag*{\rm (A)} \]

    on \((a,b)\). Show that if \(y\) is a solution of (A) on \((a,b)\), there’s exactly one way to choose \(c_1\) and \(c_2\) so that \(y=c_1y_1+c_2y_2\) on \((a,b)\).

37. Suppose \(p\) and \(q\) are continuous on \((a,b)\) and \(x_0\) is in \((a,b)\). Let \(y_1\) and \(y_2\) be the solutions of

    \[ y''+p(x)y'+q(x)y=0 \tag*{\rm (A)} \]

    such that

    \[ y_1(x_0)=1, \quad y'_1(x_0)=0\mbox{\quad and \quad} y_2(x_0)=0,\; y'_2(x_0)=1. \]

    (Theorem 5.1.1 implies that each of these initial value problems has a unique solution on \((a,b)\).)

    1. Show that \(\{y_1,y_2\}\) is linearly independent on \((a,b)\).

    2. Show that an arbitrary solution \(y\) of (A) on \((a,b)\) can be written as \(y=y(x_0)y_1+y'(x_0)y_2\).

    3. Express the solution of the initial value problem

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

       as a linear combination of \(y_1\) and \(y_2\).

    Show answer

    (c) \(y=k_0y_1+k_1y_2\)

38. Find solutions \(y_1\) and \(y_2\) of the equation \(y''=0\) that satisfy the initial conditions

    \[ y_1(x_0)=1, \quad y'_1(x_0)=0 \mbox{\quad and \quad} y_2(x_0)=0, \quad y'_2(x_0)=1. \]

    Then use Exercise 37 (c) to write the solution of the initial value problem

    \[ y''=0,\quad y(0)=k_0,\quad y'(0)=k_1 \]

    as a linear combination of \(y_1\) and \(y_2\).

    Show answer

    \(y_1=1\), \(y_2=x-x_0\); \(y=k_0+k_1(x-x_0)\)

39. Let \(x_0\) be an arbitrary real number. Given (Example 5.1.1) that \(e^x\) and \(e^{-x}\) are solutions of \(y''-y=0\), find solutions \(y_1\) and \(y_2\) of \(y''-y=0\) such that

    \[ y_1(x_0)=1, \quad y'_1(x_0)=0\mbox{\quad and \quad} y_2(x_0)=0,\; y'_2(x_0)=1. \]

    Then use Exercise 37 (c) to write the solution of the initial value problem

    \[ y''-y=0,\quad y(x_0)=k_0,\quad y'(x_0)=k_1 \]

    as a linear combination of \(y_1\) and \(y_2\).

    Show answer

    \(y_1=\cosh(x-x_0)\), \(y_2=\sinh(x-x_0)\); \(y=k_0\cosh(x-x_0)+k_1\sinh(x-x_0)\)

40. Let \(x_0\) be an arbitrary real number. Given (Example 5.1.2) that \(\cos\omega x\) and \(\sin\omega x\) are solutions of \(y''+\omega^2y=0\), find solutions of \(y''+\omega^2y=0\) such that

    \[ y_1(x_0)=1, \quad y'_1(x_0)=0\mbox{\quad and \quad} y_2(x_0)=0,\; y'_2(x_0)=1. \]

    Then use Exercise 37 (c) to write the solution of the initial value problem

    \[ y''+\omega^2y=0,\quad y(x_0)=k_0,\quad y'(x_0)=k_1 \]

    as a linear combination of \(y_1\) and \(y_2\). Use the identities

    \begin{eqnarray*} \cos(A+B)&=&\cos A\cos B-\sin A\sin B \\ \sin(A+B)&=&\sin A\cos B+\cos A\sin B \end{eqnarray*}

    to simplify your expressions for \(y_1\), \(y_2\), and \(y\).

    Show answer

    \(y_1=\cos\omega(x-x_0)\), \(y_2=\dst{1\over\omega}\sin\omega(x-x_0)\) \(y=k_0\cos\omega(x-x_0)+\dst{k_1\over\omega}\sin\omega(x-x_0)\)

41. Recall from Exercise 4 that \(1/(x-1)\) and \(1/(x+1)\) are solutions of

    \[ (x^2-1)y''+4xy'+2y=0 \tag*{\rm (A)} \]

    on \((-1,1)\). Find solutions of (A) such that

    \[ y_1(0)=1, \quad y'_1(0)=0\mbox{\quad and \quad} y_2(0)=0,\; y'_2(0)=1. \]

    Then use Exercise 37 (c) to write the solution of initial value problem

    \[ (x^2-1)y''+4xy'+2y=0,\quad y(0)=k_0,\quad y'(0)=k_1 \]

    as a linear combination of \(y_1\) and \(y_2\).

    Show answer

    \(y_1=\dst{1\over1-x^2}\),  \(y_2=\dst{x\over1-x^2}\) \(y=\dst{k_0+k_1x\over1-x^2}\)

42. 1. Verify that \(y_1=x^2\) and \(y_2=x^3\) satisfy

       \[ x^2y''-4xy'+6y=0 \tag*{\rm (A)} \]

       on \((-\infty,\infty)\) and that \(\{y_1,y_2\}\) is a fundamental set of solutions of (A) on \((-\infty,0)\) and \((0,\infty)\).

    2. Let \(a_1\), \(a_2\), \(b_1\), and \(b_2\) be constants. Show that

       \[ y=\left\{\begin{array}{rr} a_1x^2+a_2x^3,&x\ge 0,\\ b_1x^2+b_2x^3,&x<0\phantom{,} \end{array}\right. \]

       is a solution of (A) on \((-\infty,\infty)\) if and only if \(a_1=b_1\). From this, justify the statement that \(y\) is a solution of (A) on \((-\infty,\infty)\) if and only if

       \[ y=\left\{\begin{array}{rr} c_1x^2+c_2x^3,&x\ge 0,\\ c_1x^2+c_3x^3,&x<0, \end{array}\right. \]

       where \(c_1\), \(c_2\), and \(c_3\) are arbitrary constants.

    3. For what values of \(k_0\) and \(k_1\) does the initial value problem

       \[ x^2y''-4xy'+6y=0,\quad y(0)=k_0,\quad y'(0)=k_1 \]

       have a solution? What are the solutions?

    4. Show that if \(x_0\ne0\) and \(k_0,k_1\) are arbitrary constants, the initial value problem

       \[ x^2y''-4xy'+6y=0,\quad y(x_0)=k_0,\quad y'(x_0)=k_1 \tag*{\rm (B)} \]

       has infinitely many solutions on \((-\infty,\infty)\). On what interval does (B) have a unique solution?

    Show answer

    (c) \(k_0=k_1=0\); \(y=\dst{\left\{\begin{array}{rr} c_1x^2+c_2x^3,x\ge 0,\\ c_1x^2+c_3x^3,x<0 \end{array}\right.}\) 

         (d) \((0,\infty)\) if \(x_0>0\), \((-\infty,0)\) if \(x_0<0\)

43. 1. Verify that \(y_1=x\) and \(y_2=x^2\) satisfy

       \[ x^2y''-2xy'+2y=0 \tag*{\rm (A)} \]

       on \((-\infty,\infty)\) and that \(\{y_1,y_2\}\) is a fundamental set of solutions of (A) on \((-\infty,0)\) and \((0,\infty)\).

    2. Let \(a_1\), \(a_2\), \(b_1\), and \(b_2\) be constants. Show that

       \[ y=\left\{\begin{array}{rr} a_1x+a_2x^2,&x\ge 0,\\ b_1x+b_2x^2,&x<0\phantom{,} \end{array}\right. \]

       is a solution of (A) on \((-\infty,\infty)\) if and only if \(a_1=b_1\) and \(a_2=b_2\). From this, justify the statement that the general solution of (A) on \((-\infty,\infty)\) is \(y=c_1x+c_2x^2\), where \(c_1\) and \(c_2\) are arbitrary constants.

    3. For what values of \(k_0\) and \(k_1\) does the initial value problem

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

       have a solution? What are the solutions?

    4. Show that if \(x_0\ne0\) and \(k_0,k_1\) are arbitrary constants then the initial value problem

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

       has a unique solution on \((-\infty,\infty)\).

    Show answer

    (c) \(k_0=0\), \(k_1\) arbitrary  \(y=k_1x+c_2x^2\)

44. 1. Verify that \(y_1=x^3\) and \(y_2=x^4\) satisfy

       \[ x^2y''-6xy'+12y=0 \tag*{\rm (A)} \]

       on \((-\infty,\infty)\), and that \(\{y_1,y_2\}\) is a fundamental set of solutions of (A) on \((-\infty,0)\) and \((0,\infty)\).

    2. Show that \(y\) is a solution of (A) on \((-\infty,\infty)\) if and only if

       \[ y=\left\{\begin{array}{rr} a_1x^3+a_2x^4,&x\ge 0,\\ b_1x^3+b_2x^4,&x<0, \end{array}\right. \]

       where \(a_1\), \(a_2\), \(b_1\), and \(b_2\) are arbitrary constants.

    3. For what values of \(k_0\) and \(k_1\) does the initial value problem

       \[ x^2y''-6xy'+12y=0, \quad y(0)=k_0,\quad y'(0)=k_1 \]

       have a solution? What are the solutions?

    4. Show that if \(x_0\ne0\) and \(k_0,k_1\) are arbitrary constants then the initial value problem

       \[ x^2y''-6xy'+12y=0, \quad y(x_0)=k_0,\quad y'(x_0)=k_1 \tag*{\rm (B)} \]

       has infinitely many solutions on \((-\infty,\infty)\). On what interval does (B) have a unique solution?

    Show answer

    (c) \(k_0=k_1=0\) \(y=\dst{\left\{\begin{array}{rr} a_1x^3+a_2x^4,x\ge 0,\\ b_1x^3+b_2x^4,x<0 \end{array}\right.}\) 

         (d) \((0,\infty)\) if \(x_0>0\), \((-\infty,0)\) if \(x_0<0\)