1. Auxiliary Equation with Complex Roots; Hyperbolic Functions
MATH 374 Lecture 17
Auxiliary Equation with Complex Roots; Hyperbolic Functions

2. 5.5: The Auxiliary Equation: Complex Roots
Given the nth order linear homogeneous differential equation with constant coefficients,
f(D)y = 0 (1)
suppose the auxiliary equation f(m) = 0 has real coefficients and
m1 = a + i b (a, b real with b  0)
is a root. 2

3. Auxiliary Equation with Complex Roots
f(D)y = 0 (1)
Auxiliary Equation with Complex Roots
Then, by Remark 4 in our Section 5.3 & 5.4 notes,
m2 = a - i b is also a root of f(m) = 0.
It follows that
f(D) = g(D)(D – (a+ib))(D – (a-ib)),
so from Example 2 in our Section 5.3 & 5.4 notes, we see that
y = c1 e(a+ib)x + c2 e(a-ib)x (2)
is a solution of (1). 3

4. Auxiliary Equation with Complex Roots
f(D)y = 0 (1)
Auxiliary Equation with Complex Roots
y = c1 e(a+ib)x + c2 e(a-ib)x (2)
Using (7) in our Section 5.3 & 5.4 notes, we can rewrite (2):
y = c1eax(cos bx + i sin bx) + c2eax(cos bx – i sin bx)
= (c1+c2)eax cos bx + i(c1 – c2)eax sin bx.
Letting c3 = c1 + c2 and c4 = i(c1 – c2), we see that
y = c3 eax cos bx + c4eax sin bx (3)
is a solution of (1) corresponding to the roots a § i b of f(m) = 0.
Note: eax cos bx and eax sin bx are linearly independent! 4

5. Auxiliary Equation with Complex Roots
f(D)y = 0 (1)
Auxiliary Equation with Complex Roots
What we’ve just shown is:
Theorem 5.4: If for the linear homogeneous equation (1), the auxiliary equation f(m) = 0 has real coefficients and m1 = a+ib (a, b real and b  0) is a root, then so is m2 = a-ib and
y = c1 eax cos bx + c2eax sin bx
where c1 and c2 are constants is a solution to (1).
Remark: For repeated complex or real roots, a theorem similar to Theorem 5.2 holds!

6. Example 1: Find the general solution of y’’’ + y’’ + 4y’ + 4y = 0.
Solution: The auxiliary equation is
m3 + m2 + 4m + 4 = 0. -1 1 4

7. Example 1: Find the general solution of y’’’ + y’’ + 4y’ + 4y = 0.
Solution: The auxiliary equation is
m3 + m2 + 4m + 4 = 0. -1 1 4

8. Example 1: Find the general solution of y’’’ + y’’ + 4y’ + 4y = 0.
Solution: The auxiliary equation is
m3 + m2 + 4m + 4 = 0. -1 1 4

9. Example 1: Find the general solution of y’’’ + y’’ + 4y’ + 4y = 0.
Solution: The auxiliary equation is
m3 + m2 + 4m + 4 = 0. -1 1 4

10. Example 1: Find the general solution of y’’’ + y’’ + 4y’ + 4y = 0.
Solution: The auxiliary equation is
m3 + m2 + 4m + 4 = 0. -1 1 4

11. Example 1: Find the general solution of y’’’ + y’’ + 4y’ + 4y = 0.
Solution: The auxiliary equation is
m3 + m2 + 4m + 4 = 0. -1 1 4

12. Example 1: Find the general solution of y’’’ + y’’ + 4y’ + 4y = 0.
Solution: The auxiliary equation is
m3 + m2 + 4m + 4 = 0. -1 1 4 -4

13. Example 1: Find the general solution of y’’’ + y’’ + 4y’ + 4y = 0.
Solution: The auxiliary equation is
m3 + m2 + 4m + 4 = 0. -1 1 4 -4
) (m+1)(m2 + 4) =0
) m = -1, § 2i
) y = c1 e-x + c2 e0·x cos 2x + c3 e0·x sin 2x
= c1 e-x + c2 cos 2x + c3 sin 2x

14. Example 2: Solve [(D-5)2 + 9]2 y = 0
Solution: Note that for the auxiliary equation
[(m-5)2 + 9]2 = 0,
the roots are m = 5 § 3i, 5 § 3i.
Hence,
y = c1 e5x cos 3x + c2 x e5x cos 3x c3 e5x sin 3x + c4 x e5x sin 3x.
Note: In general, for a,b real, a § bi are roots of (m-a)2 + b2 = 0.

15. 5.6: Hyperbolic Functions
Definition: The hyperbolic sine and hyperbolic cosine are the functions
y = sinh x
y = cosh x

16. Properties of sinh x and cosh x
cosh2 x – sinh2 x = 1
d/dx[cosh x] = sinh x and d/dx[sinh x] = cosh x.
cosh x is even. sinh x is odd.
y1 = cosh(ax) and y2 = sinh(ax) are linearly independent solutions of
(D2 – a2)y = 0. (a real, a  0) (1)

17. Properties of sinh x and cosh x
y1 = cosh(ax) and y2 = sinh(ax) are linearly independent solutions of
(D2 – a2)y = 0. (a real, a  0) (1)
Properties of sinh x and cosh x
Note: From property 4, it follows that solutions of (1) can be written in the form
y = c1 cosh(ax) + c2 sinh(ax) (2)
instead of
y = c1 eax + c2 e-ax (3)
as before.
Solution form (2) is useful when solving IVP with initial data taken at x = 0.
From the definition of cosh x and sinh x, it is clear that (2) and (3) are equivalent! (check)