- #1

Dao Tuat

- 16

- 0

Heres what I was given:

4y'' - 4y' - 3y=0

y(0)=1

y'(0)=5

Heres what I did:

(sorry about the c1 and c2, I don't know how to make them show up as subsript)

4(r^2) - 4r - 3 = 0

r = 3/2 r=-1/2

y = c1e^(3x/2) + c2e^(-x/2)

1 = c1 + c2

y'' = [3c1e^(3x/2)]/2 - [c2e^(-x/2)]/2

5 = 3c1/2 - c2/2

c2 = ±2sqrt(7) - 5

c1 = 1/(±2sqrt(7) - 5) = [±2sqrt(7)]/3 + 5/3

y = [(±2sqrt(7))/3 + 5/3]e(3x/2) + [±2sqrt(7) - 5]e(-x/2)

Does this seem at all right?