Solving a First Order Initial Value Problem

  • Thread starter KevinD6
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  • #1
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Homework Statement


If y = y(t) is the solution of the initial value problem
y' + (2 t + 1) y = 2 cos(t)
y(0) = 2
What is y''(0)?

Homework Equations




The Attempt at a Solution


Since this is a first order linear, I started out by finding the integrating factor so I can find what y is, and then just take the second derivative of it, and then put it to 0.

[itex] IF = e^{ \int {2t + 1} } => IF = e^{t^2 + t} [/itex]
Then, I end up with this equation, if we set IF = k.
[itex] (ky)' = k (2 cos(t)) [/itex]

From here, I don't think I can integrate that equation, so now I'm pretty much stuck. Is there a method I'm missing? Or could I find the value of the second derivative using the value of y at 0?
Any help is appreciated, thank you.
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement


If y = y(t) is the solution of the initial value problem
y' + (2 t + 1) y = 2 cos(t)
y(0) = 2
What is y''(0)?

Homework Equations




The Attempt at a Solution


Since this is a first order linear, I started out by finding the integrating factor so I can find what y is, and then just take the second derivative of it, and then put it to 0.

[itex] IF = e^{ \int {2t + 1} } => IF = e^{t^2 + t} [/itex]
Then, I end up with this equation, if we set IF = k.
[itex] (ky)' = k (2 cos(t)) [/itex]

From here, I don't think I can integrate that equation, so now I'm pretty much stuck. Is there a method I'm missing? Or could I find the value of the second derivative using the value of y at 0?
Any help is appreciated, thank you.
Instead of trying to solve the original DE, try just differentiating it once and solving for ##y''## and see if you can get the answer from that.
 
  • #3
10
0
Wow! I can't believe I didn't see that, haha, thanks man. Much appreciated.
 

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