Second Order DE

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Homework Statement



[tex]y'' - 4y = 0[/tex] when y = 1, y' = -1, x = 0

2. The attempt at a solution

[tex]y'' - 4y = 0[/tex]

[tex]m^2 - 4 = 0[/tex]

[tex]m = 2, m = -2[/tex]

Substituting: [tex]y = 1, x = 0[/tex]

[tex]1 = C1 + C2[/tex]

[tex]C1 = -C2 + 1[/tex]

Substituting: [tex]y' = -1, x = 0[/tex]

[tex]-1 -C2 + 1 + C2[/tex]

[tex]C2 = C2[/tex]

Therefore: [tex]y = -C2e^2^x + C2e^-^2^x[/tex]

Obviously I went wrong with my substitutions... Can I get a hint?
 

Answers and Replies

  • #2
Try solving the initial value as a system of equations, maybe this will give you a better approach than substitution.
 
  • #3
since m=+/-2
y is of the form
y=A(e^-2x)+B(e^2x)
y`=-2A(e^-2x)+2B(e^2x)
Use the given conditions. Make two equations... You know the rest.
Hope this helps.
Abdullah

Whats wrong with the forums? It wont allow me to start a new thread. It keeps saying:

The page your looking for can't be found.
Please go back to the homepage: https://www.physicsforums.com/
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  • #4
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thanks guys, I got it
 
  • #5
HallsofIvy
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Homework Statement



[tex]y'' - 4y = 0[/tex] when y = 1, y' = -1, x = 0

2. The attempt at a solution

[tex]y'' - 4y = 0[/tex]

[tex]m^2 - 4 = 0[/tex]

[tex]m = 2, m = -2[/tex]
It would be a really smart idea to write out the general solution here:
y= C1e2x+ C2e-2x
Otherwise no one knows what the rest of this means!

Substituting: [tex]y = 1, x = 0[/tex]

[tex]1 = C1 + C2[/tex]
Okay, C1e0+ C2e0= C1+ C2= 1

[tex]C1 = -C2 + 1[/tex]

Substituting: [tex]y' = -1, x = 0[/tex]

[tex]-1 -C2 + 1 + C2[/tex]
Isn't there an "=" missing here?
If y(x)= C1e2x+ C2e-2x
then y'(x)= 2C1e2x- 2C2e-2x
so y'(0)= 2C1- 2C2= -1

[tex]C2 = C2[/tex]
Surely that's not what you meant to say!

Therefore: [tex]y = -C2e^2^x + C2e^-^2^x[/tex]

Obviously I went wrong with my substitutions... Can I get a hint?
You have C1+ C2= 1 and 2C1- 2C2= -1. Solve those two equations.
 
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