5.2 Constant Coefficient Homogeneous Equations

If \(a,b\), and \(c\) are real constants and \(a\ne0\), then

is said to be a constant coefficient equation. In this section we consider the homogeneous constant coefficient equation

As we’ll see, all solutions of (5.2.1) are defined on \((-\infty,\infty)\). This being the case, we’ll omit references to the interval on which solutions are defined, or on which a given set of solutions is a fundamental set, etc., since the interval will always be \((-\infty,\infty)\).

The key to solving (5.2.1) is that if \(y=e^{rx}\) where \(r\) is a constant then the left side of (5.2.1) is a multiple of \(e^{rx}\); thus, if \(y=e^{rx}\) then \(y'=re^{rx}\) and \(y''=r^2e^{rx}\), so

The quadratic polynomial

is the characteristic polynomial of (5.2.1), and \(p(r)=0\) is the characteristic equation. From (5.2.2) we can see that \(y=e^{rx}\) is a solution of (5.2.1) if and only if \(p(r)=0\).

The roots of the characteristic equation are given by the quadratic formula

We consider three cases:

• Case 1. \(b^2-4ac>0\), so the characteristic equation has two distinct real roots.

• Case 2. \(b^2-4ac=0\), so the characteristic equation has a repeated real root.

• Case 3. \(b^2-4ac<0\), so the characteristic equation has complex roots.

In each case we’ll start with an example.

Case 1: Distinct Real Roots

Solution (a) The characteristic polynomial of (5.2.4) is

Since \(p(-1)=p(-5)=0\), \(y_1=e^{-x}\) and \(y_2=e^{-5x}\) are solutions of (5.2.4). Since \(y_2/y_1=e^{-4x}\) is nonconstant, 5.1.6 implies that the general solution of (5.2.4) is

Solution (b) We must determine \(c_1\) and \(c_2\) in (5.2.6) so that \(y\) satisfies the initial conditions in (5.2.5). Differentiating (5.2.6) yields

Imposing the initial conditions \(y(0)=3,\, y'(0)=-1\) in (5.2.6) and (5.2.7) yields

The solution of this system is \(c_1=7/2,c_2=-1/2\). Therefore the solution of (5.2.5) is

Figure 5.2.1 is a graph of this solution.

If the characteristic equation has arbitrary distinct real roots \(r_1\) and \(r_2\), then \(y_1=e^{r_1x}\) and \(y_2=e^{r_2x}\) are solutions of \(ay''+by'+cy=0\). Since \(y_2/y_1=e^{(r_2-r_1)x}\) is nonconstant, Theorem 5.1.6 implies that \(\{y_1,y_2\}\) is a fundamental set of solutions of \(ay''+by'+cy=0\).

Case 2: A Repeated Real Root

Solution (a) The characteristic polynomial of (5.2.8) is

so the characteristic equation has the repeated real root \(r_1=-3\). Therefore \(y_1=e^{-3x}\) is a solution of (5.2.8). Since the characteristic equation has no other roots, (5.2.8) has no other solutions of the form \(e^{rx}\). We look for solutions of the form \(y=uy_1=ue^{-3x}\), where \(u\) is a function that we’ll now determine. (This should remind you of the method of variation of parameters used in Section 2.1 to solve the nonhomogeneous equation \(y'+p(x)y=f(x)\), given a solution \(y_1\) of the complementary equation \(y'+p(x)y=0\). It’s also a special case of a method called reduction of order that we’ll study in Section 5.6. For other ways to obtain a second solution of (5.2.8) that’s not a multiple of \(e^{-3x}\), see Exercises 5.1. 9, 5.1. 12, and 33.

If \(y=ue^{-3x}\), then

so

Therefore \(y=ue^{-3x}\) is a solution of (5.2.8) if and only if \(u''=0\), which is equivalent to \(u=c_1+c_2x\), where \(c_1\) and \(c_2\) are constants. Therefore any function of the form

is a solution of (5.2.8). Letting \(c_1=1\) and \(c_2=0\) yields the solution \(y_1=e^{-3x}\) that we already knew. Letting \(c_1=0\) and \(c_2=1\) yields the second solution \(y_2=xe^{-3x}\). Since \(y_2/y_1=x\) is nonconstant, 5.1.6 implies that \(\{y_1,y_2\}\) is fundamental set of solutions of (5.2.8), and (5.2.10) is the general solution.

Solution (b) Differentiating (5.2.10) yields

Imposing the initial conditions \(y(0)=3,\, y'(0)=-1\) in (5.2.10) and (5.2.11) yields \(c_1=3\) and \(-3c_1+c_2=-1\), so \(c_2=8\). Therefore the solution of (5.2.9) is

Figure 5.2.2 is a graph of this solution.

If the characteristic equation of \(ay''+by'+cy=0\) has an arbitrary repeated root \(r_1\), the characteristic polynomial must be

Therefore

which implies that \(b=-2ar_1\) and \(c=ar_1^2\). Therefore \(ay''+by'+cy=0\) can be written as \(a(y''-2r_1y'+r_1^2y)=0\). Since \(a\ne0\) this equation has the same solutions as

Since \(p(r_1)=0\), t \(y_1=e^{r_1x}\) is a solution of \(ay''+by'+cy=0\), and therefore of (5.2.12). Proceeding as in Example 5.2.2, we look for other solutions of (5.2.12) of the form \(y=ue^{r_1x}\); then

so

Therefore \(y=ue^{r_1x}\) is a solution of (5.2.12) if and only if \(u''=0\), which is equivalent to \(u=c_1+c_2x\), where \(c_1\) and \(c_2\) are constants. Hence, any function of the form

is a solution of (5.2.12). Letting \(c_1=1\) and \(c_2=0\) here yields the solution \(y_1=e^{r_1x}\) that we already knew. Letting \(c_1=0\) and \(c_2=1\) yields the second solution \(y_2=xe^{r_1x}\). Since \(y_2/y_1=x\) is nonconstant, 5.1.6 implies that \(\{y_1,y_2\}\) is a fundamental set of solutions of (5.2.12), and (5.2.13) is the general solution.

Case 3: Complex Conjugate Roots

Solution (a) The characteristic polynomial of (5.2.14) is

The roots of the characteristic equation are \(r_1=-2+3i\) and \(r_2=-2-3i\). By analogy with Case 1, it’s reasonable to expect that \(e^{(-2+3i)x}\) and \(e^{(-2-3i)x}\) are solutions of (5.2.14). This is true (see Exercise 34); however, there are difficulties here, since you are probably not familiar with exponential functions with complex arguments, and even if you are, it’s inconvenient to work with them, since they are complex–valued. We’ll take a simpler approach, which we motivate as follows: the exponential notation suggests that

so even though we haven’t defined \(e^{3ix}\) and \(e^{-3ix}\), it’s reasonable to expect that every linear combination of \(e^{(-2+3i)x}\) and \(e^{(-2-3i)x}\) can be written as \(y=ue^{-2x}\), where \(u\) depends upon \(x\). To determine \(u\), we note that if \(y=ue^{-2x}\) then

so

Therefore \(y=ue^{-2x}\) is a solution of (5.2.14) if and only if

From Example 5.1.2, the general solution of this equation is

Therefore any function of the form

is a solution of (5.2.14). Letting \(c_1=1\) and \(c_2=0\) yields the solution \(y_1=e^{-2x}\cos3x\). Letting \(c_1=0\) and \(c_2=1\) yields the second solution \(y_2=e^{-2x}\sin3x\). Since \(y_2/y_1=\tan3x\) is nonconstant, 5.1.6 implies that \(\{y_1,y_2\}\) is a fundamental set of solutions of (5.2.14), and (5.2.16) is the general solution.

Solution (b) Imposing the condition \(y(0)=2\) in (5.2.16) shows that \(c_1=2\). Differentiating (5.2.16) yields

and imposing the initial condition \(y'(0)=-3\) here yields \(-3=-2c_1+3c_2=-4+3c_2\), so \(c_2=1/3\). Therefore the solution of (5.2.15) is

Figure 5.2.3 is a graph of this function. ∎

Now suppose the characteristic equation of \(ay''+by'+cy=0\) has arbitrary complex roots; thus, \(b^2-4ac<0\) and, from (5.2.3), the roots are

which we rewrite as

with

Don’t memorize these formulas. Just remember that \(r_1\) and \(r_2\) are of the form (5.2.17), where \(\lambda\) is an arbitrary real number and \(\omega\) is positive; \(\lambda\) and \(\omega\) are the real and imaginary parts, respectively, of \(r_1\). Similarly, \(\lambda\) and \(-\omega\) are the real and imaginary parts of \(r_2\). We say that \(r_1\) and \(r_2\) are complex conjugates, which means that they have the same real part and their imaginary parts have the same absolute values, but opposite signs.

As in Example 5.2.3, it’s reasonable to to expect that the solutions of \(ay''+by'+cy=0\) are linear combinations of \(e^{(\lambda+i\omega)x}\) and \(e^{(\lambda-i\omega)x}\). Again, the exponential notation suggests that

so even though we haven’t defined \(e^{i\omega x}\) and \(e^{-i\omega x}\), it’s reasonable to expect that every linear combination of \(e^{(\lambda+i\omega)x}\) and \(e^{(\lambda-i\omega)x}\) can be written as \(y=ue^{\lambda x}\), where \(u\) depends upon \(x\). To determine \(u\) we first observe that since \(r_1=\lambda+i\omega\) and \(r_2=\lambda-i\omega\) are the roots of the characteristic equation, \(p\) must be of the form

Therefore \(ay''+by'+cy=0\) can be written as

Since \(a\ne0\) this equation has the same solutions as

To determine \(u\) we note that if \(y=ue^{\lambda x}\) then

Substituting these expressions into (5.2.18) and dropping the common factor \(e^{\lambda x}\) yields

which simplifies to

From Example 5.1.2, the general solution of this equation is

Therefore any function of the form

is a solution of (5.2.18). Letting \(c_1=1\) and \(c_2=0\) here yields the solution \(y_1=e^{\lambda x}\cos\omega x\). Letting \(c_1=0\) and \(c_2=1\) yields a second solution \(y_2=e^{\lambda x}\sin\omega x\). Since \(y_2/y_1=\tan\omega x\) is nonconstant, so Theorem 5.1.6 implies that \(\{y_1,y_2\}\) is a fundamental set of solutions of (5.2.18), and (5.2.19) is the general solution.

Summary

The next theorem summarizes the results of this section.

5.2 Exercises

In Exercises 1–12 find the general solution.

1. \(y''+5y'-6y=0\)

   Show answer

   \(y=c_1e^{-6x}+c_2e^{x}\)

2. \(y''-4y'+5y=0\)

   Show answer

   \(y=e^{2x}(c_1 \cos x+c_2 \sin x)\)

3. \(y''+8y'+7y=0\)

   Show answer

   \(y=c_1e^{-7x}+c_2e^{-x}\)

4. \(y''-4y'+4y=0\)

   Show answer

   \(y=e^{2x}(c_1+c_2x)\)

5. \(y''+2y'+10y=0\)

   Show answer

   \(y=e^{-x}(c_1 \cos 3x+c_2 \sin 3x)\)

6. \(y''+6y'+10y=0\)

   Show answer

   \(y=e^{-3x}(c_1 \cos x+c_2 \sin x)\)

7. \(y''-8y'+16y=0\)

   Show answer

   \(y=e^{4x}(c_1+c_2x)\)

8. \(y''+y'=0\)

   Show answer

   \(y=c_1+c_2e^{-x}\)

9. \(y''-2y'+3y=0\)

   Show answer

   \(y=e^x(c_1 \cos \sqrt{2}x+c_2 \sin \sqrt{2} x)\)

10. \(y''+6y'+13y=0 \)

    Show answer

    \(y=e^{-3x}(c_1\cos 2x+c_2\sin 2x)\)

11. \(4y''+4y'+10y=0\)

    Show answer

    \(\dst{y=e^{-x/2}\left(c_1\cos {3x\over2}+c_2\sin {3x\over2}\right)}\)

12. \(10y''-3y'-y=0\)

    Show answer

    \(y=c_1e^{-x/5}+c_2e^{x/2}\)

13. \(y''+14y'+50y=0, \quad y(0)=2,\quad y'(0)=-17\)

    Show answer

    \(y=e^{-7x}(2 \cos x - 3 \sin x)\)

14. \(6y''-y'-y=0, \quad y(0)=10,\quad y'(0)=0\)

    Show answer

    \(y=4e^{x/2}+6e^{-x/3}\)

15. \(6y''+y'-y=0, \quad y(0)=-1,\quad y'(0)=3\)

    Show answer

    \(y=3e^{x/3}-4e^{-x/2}\)

16. \(4y''-4y'-3y=0, \quad y(0)=\dst{13\over 12},\quad y'(0)=\dst{23 \over 24}\)

    Show answer

    \(y=\dst{e^{-x/2}\over 3}+\dst{3e^{3x/2}\over4}\)

17. \(4y''-12y'+9y=0, \quad y(0)=3,\quad y'(0)=\dst{5\over 2}\)

    Show answer

    \(y=e^{3x/2} (3-2x)\)

18. C/G \(y''+7y'+12y=0, \quad y(0)=-1,\quad y'(0)=0\)

    Show answer

    \(y=3e^{-4x}-4e^{-3x}\)

19. C/G \(y''-6y'+9y=0, \quad y(0)=0,\quad y'(0)=2\)

    Show answer

    \(y=2xe^{3x}\)

20. C/G \(36y''-12y'+y=0, \quad y(0)=3,\quad y'(0)=\dst{5\over2}\)

    Show answer

    \(y=e^{x/6}(3+2x)\)

21. C/G \(y''+4y'+10y=0, \quad y(0)=3,\quad y'(0)=-2\)

    Show answer

    \(\dst{y=e^{-2x}\left(3 \cos \sqrt{6} x+{2 \sqrt{6} \over 3} \sin \sqrt{6} x\right)}\)

22. 1. Suppose \(y\) is a solution of the constant coefficient homogeneous equation

       \[ ay''+by'+cy=0. \tag*{\rm (A)} \]

       Let \(z(x)=y(x-x_0)\), where \(x_0\) is an arbitrary real number. Show that

       \[ az''+bz'+cz=0. \]

    2. Let \(z_1(x)=y_1(x-x_0)\) and \(z_2(x)=y_2(x-x_0)\), where \(\{y_1,y_2\}\) is a fundamental set of solutions of (A). Show that \(\{z_1,z_2\}\) is also a fundamental set of solutions of (A).

    3. The statement of Theorem 5.2.1 is convenient for solving an initial value problem

       \[ ay''+by'+cy=0, \quad y(0)=k_0,\quad y'(0)=k_1, \]

       where the initial conditions are imposed at \(x_0=0\). However, if the initial value problem is

       \[ ay''+by'+cy=0, \quad y(x_0)=k_0,\quad y'(x_0)=k_1, \tag*{\rm (B)} \]

       where \(x_0\ne0\), then determining the constants in

       \[ y=c_1e^{r_1x}+c_2e^{r_2x}, \quad y=e^{r_1x}(c_1+c_2x),\mbox{ or } y=e^{\lambda x}(c_1\cos\omega x+c_2\sin\omega x) \]

       (whichever is applicable) is more complicated. Use (b) to restate Theorem 5.2.1 in a form more convenient for solving (B).

23. \(y''+3y'+2y=0, \quad y(1)=-1,\quad y'(1)=4\)

    Show answer

    \(y=2e^{-(x-1)}-3e^{-2(x-1)}\)

24. \(y''-6y'-7y=0, \quad y(2)=-\dst{1\over3},\quad y'(2)=-5\)

    Show answer

    \(y=\dst{{1\over3}e^{-(x-2)}-{2\over3}e^{7(x-2)}}\)

25. \(y''-14y'+49y=0, \quad y(1)=2,\quad y'(1)=11\)

    Show answer

    \(y=e^{7(x-1)}\left(2-3(x-1)\right)\)

26. \(9y''+6y'+y=0, \quad y(2)=2,\quad y'(2)=-\dst{14\over3}\)

    Show answer

    \(y=e^{-(x-2)/3}\left(2-4(x-2)\right)\)

27. \(9y''+4y=0, \quad y(\pi/4)=2,\quad y'(\pi/4)=-2\)

    Show answer

    \(y=\dst{2\cos {2\over3}\left(x-{\pi\over4}\right)-3\sin {2\over3}\left(x-{\pi\over4}\right)}\)

28. \(y''+3y=0, \quad y(\pi/3)=2,\quad y'(\pi/3)=-1\)

    Show answer

    \(y=\dst{2\cos \sqrt3\left(x-{\pi\over3}\right)-{1\over\sqrt3}\sin \sqrt3\left(x-{\pi\over3}\right)}\)

29. Prove: If the characteristic equation of

    \[ ay''+by'+cy=0 \tag*{\rm (A)} \]

    has a repeated negative root or two roots with negative real parts, then every solution of (A) approaches zero as \(x\to\infty\).

30. Suppose the characteristic polynomial of \(ay''+by'+cy=0\) has distinct real roots \(r_1\) and \(r_2\). Use a method suggested by Exercise 22 to find a formula for the solution of

    \[ ay''+by'+cy=0, \quad y(x_0)=k_0,\quad y'(x_0)=k_1. \]
    Show answer

    \(y=\dst{{k_0\over r_2-r_1}\left(r_2e^{r_1(x-x_0)}-r_1e^{r_2(x-x_0)}\right)+{k_1\over r_2-r_1} \left(e^{r_2(x-x_0)}-e^{r_1(x-x_0)}\right)}\)

31. Suppose the characteristic polynomial of \(ay''+by'+cy=0\) has a repeated real root \(r_1\). Use a method suggested by Exercise 22 to find a formula for the solution of

    \[ ay''+by'+cy=0, \quad y(x_0)=k_0,\quad y'(x_0)=k_1. \]
    Show answer

    \(y=e^{r_1(x-x_0)}\left[k_0+(k_1-r_1k_0)(x-x_0)\right]\)

32. Suppose the characteristic polynomial of \(ay''+by'+cy=0\) has complex conjugate roots \(\lambda\pm i\omega\). Use a method suggested by Exercise 22 to find a formula for the solution of

    \[ ay''+by'+cy=0, \quad y(x_0)=k_0,\quad y'(x_0)=k_1. \]
    Show answer

    \(y=\dst{e^{\lambda(x-x_0)}\left[k_0\cos\omega(x-x_0)+\left(k_1-\lambda k_0 \over\omega\right)\sin\omega(x-x_0)\right]}\)

33. Suppose the characteristic equation of

    \[ ay''+by'+cy=0 \tag*{\rm (A)} \]

    has a repeated real root \(r_1\). Temporarily, think of \(e^{rx}\) as a function of two real variables \(x\) and \(r\).

    1. Show that

       \[ a{\partial^2\over\partial^2 x}(e^{rx})+b{\partial \over\partial x}(e^{rx}) +ce^{rx}=a(r-r_1)^2e^{rx}. \tag*{\rm (B)} \]

    2. Differentiate (B) with respect to \(r\) to obtain

       \[ a{\partial\over\partial r}\left({\partial^2\over\partial^2 x}(e^{rx})\right)+b{\partial\over\partial r}\left({\partial \over\partial x}(e^{rx})\right) +c(xe^{rx})=[2+(r-r_1)x]a(r-r_1)e^{rx}. \tag*{\rm (C)} \]

    3. Reverse the orders of the partial differentiations in the first two terms on the left side of (C) to obtain

       \[ a{\partial^2\over\partial x^2}(xe^{rx})+b{\partial\over\partial x}(xe^{rx})+c(xe^{rx})=[2+(r-r_1)x]a(r-r_1)e^{rx}. \tag*{\rm (D)} \]

    4. Set \(r=r_1\) in (B) and (D) to see that \(y_1=e^{r_1x}\) and \(y_2=xe^{r_1x}\) are solutions of (A)

34. In calculus you learned that \(e^u\), \(\cos u\), and \(\sin u\) can be represented by the infinite series

    \[ e^u=\sum_{n=0}^\infty {u^n\over n!} =1+{u\over 1!}+{u^2\over 2!}+{u^3\over 3!}+\cdots+{u^n\over n!}+\cdots \tag*{\rm (A)} \]
    \[ \cos u=\sum_{n=0}^\infty (-1)^n{u^{2n}\over(2n)!} =1-{u^2\over2!}+{u^4\over4!}+\cdots+(-1)^n{u^{2n}\over(2n)!} +\cdots, \tag*{\rm (B)} \]

    and

    \[ \sin u=\sum_{n=0}^\infty (-1)^n{u^{2n+1}\over(2n+1)!} =u-{u^3\over3!}+{u^5\over5!}+\cdots+(-1)^n {u^{2n+1}\over(2n+1)!} +\cdots \tag*{\rm (C)} \]

    for all real values of \(u\). Even though you have previously considered (A) only for real values of \(u\), we can set \(u=i\theta\), where \(\theta\) is real, to obtain

    \[ e^{i\theta}=\sum_{n=0}^\infty {(i\theta)^n\over n!}. \tag*{\rm (D)} \]

    Given the proper background in the theory of infinite series with complex terms, it can be shown that the series in (D) converges for all real \(\theta\).

    1. Recalling that \(i^2=-1,\) write enough terms of the sequence \(\{i^n\}\) to convince yourself that the sequence is repetitive:

       \[ 1,i,-1,-i,1,i,-1,-i,1,i,-1,-i,1,i,-1,-i,\cdots. \]

       Use this to group the terms in (D) as

       \begin{eqnarray*} e^{i\theta}&=&\left(1-{\theta^2\over2}+{\theta^4\over4}+\cdots\right) +i\left(\theta-{\theta^3\over3!}+{\theta^5\over5!}+\cdots\right) \\[6pt] &=&\sum_{n=0}^\infty (-1)^n{\theta^{2n}\over(2n)!} +i\sum_{n=0}^\infty (-1)^n{\theta^{2n+1}\over(2n+1)!}. \end{eqnarray*}

       By comparing this result with (B) and (C), conclude that

       \[ e^{i\theta}=\cos\theta+i\sin\theta. \tag*{\rm (E)} \]

       This is Euler’s identity.

    2. Starting from

       \[ e^{i\theta_1}e^{i\theta_2}=(\cos\theta_1+i\sin\theta_1) (\cos\theta_2+i\sin\theta_2), \]

       collect the real part (the terms not multiplied by \(i\)) and the imaginary part (the terms multiplied by \(i\)) on the right, and use the trigonometric identities

       \begin{eqnarray*} \cos(\theta_1+\theta_2)&=&\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2 \\ \sin(\theta_1+\theta_2)&=&\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2 \end{eqnarray*}

       to verify that

       \[ e^{i(\theta_1+\theta_2)}=e^{i\theta_1}e^{i\theta_2}, \]

       as you would expect from the use of the exponential notation \(e^{i\theta}\).

    3. If \(\alpha\) and \(\beta\) are real numbers, define

       \[ e^{\alpha+i\beta}=e^\alpha e^{i\beta}=e^\alpha(\cos\beta+i\sin\beta). \tag*{\rm (F)} \]

       Show that if \(z_1=\alpha_1+i\beta_1\) and \(z_2=\alpha_2+i\beta_2\) then

       \[ e^{z_1+z_2}=e^{z_1}e^{z_2}. \]

    4. Let \(a\), \(b\), and \(c\) be real numbers, with \(a\ne0\). Let \(z=u+iv\) where \(u\) and \(v\) are real-valued functions of \(x\). Then we say that \(z\) is a solution of

       \[ ay''+by'+cy=0 \tag*{\rm (G)} \]

       if \(u\) and \(v\) are both solutions of (G). Use Theorem 5.2.1(c) to verify that if the characteristic equation of (G) has complex conjugate roots \(\lambda\pm i\omega\) then \(z_1=e^{(\lambda+i\omega)x}\) and \(z_2=e^{(\lambda-i\omega)x}\) are both solutions of (G).