183_pdfsam_math 54 differential equation solutions odd -...

Exercises 4.3
So, α = − 1, β = 2, and a general solution is given by
y ( t ) = c
1
e
− t
cos 2 t + c
2
e
− t
sin 2 t.
15. First, we find the roots of the auxiliary equation.
r
2
+ 10 r + 41 = 0 ⇒ r =
− 10 ± 10 2 − 4(1)(41)
2
= − 5 ± 4 i.
These are complex numbers with α = − 5 and β = 4. Hence, a general solution to the given
differential equation is
y ( t ) = c
1
e
− 5 t
cos 4 t + c
2
e
− 5 t
sin 4 t.
17. The auxiliary equation in this problem, r 2 − r + 7 = 0, has the roots
r =
1 ± 1
2
− 4(1)(7)
2
=
1 ±
√
− 27
2
=
1
2
±
3
√
3
2
i.
Therefore, a general solution is
y ( t ) = c
1
e
t/ 2
cos
3
√
3
2
t + c
2
e
t/ 2
sin
3
√
3
2
t .
19. The auxiliary equation, r 3 + r 2 +3 r − 5 = 0, is a cubic equation. Since any cubic equation has
a real root, first we examine the divisors of the free coefficient, 5, to find integer real roots (if