Spring 2019
Assignment Section 3.1 Homogeneous Constant Coefficients
Answer(s) submitted:
•c1exp(2x)+c2exp(-2x)
(correct)
Correct Answers:
•c1*eˆ(2*x)+c2*eˆ(-2*x)
4. (1 point)
Find the solution to the boundary value problem:
d2y dy
−6+8y = 0, y(0)= 5,y(1)= 5 dt2
dt
y =
Solution: The charateristic equation r2−6r+8 = 0 has roots r1
= 2 and r2 = 4 so the general solution is
y = c1e2t +c2e4t
The initial conditions yield the system
5 = c1+c2
5 = c1e2+c2e4
4
Solving the system yields c1 e −eand c2e −e
Answer(s) submitted:
•-0.67667exp(4t)+5.67667exp(2t)
(correct)
Correct Answers:
•5.67667641618306*exp(2*t) + -0.676676416183064*exp(4*t)
5. (1 point)
Find y as a function of t if
y00 −9y = 0
with y(0)= 6, y0(0)= 7.
y =
Solution: The characteristic
has roots r1 = and r
equation 1r2−9 = 0
so the general solution is
y .
Substituting the initial condition gives the system
6 = c1+c2
with solution c and c.
Thus the solution of the initial value problem is
y .
Answer(s) submitted:
•((25/6)exp(3t))+((11/6)exp(-3t))
(correct)
Correct
Answers:
•(3 -
7/6)
*exp((0 - 3)*t) + (3 + 7/6) *exp((0 + 3)*t)
