How to solve the given differential equation

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SUMMARY

The discussion centers on solving the differential equation 4y'' - 4y' - 3y = 0 with initial conditions y(0) = 1 and y'(0) = 5. The user correctly identifies the characteristic equation 4r^2 - 4r - 3 = 0, yielding roots r = 3/2 and r = -1/2. The general solution is expressed as y = c1e^(3x/2) + c2e^(-x/2). The user derives the constants c1 and c2 using the initial conditions, leading to the final solution involving these constants.

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Dao Tuat
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Can someone give me some help, or at least tell me if I'm on the right track with this?

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?
 
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Looks good to me!
 

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