A system of 1st order diffy q's

  • Thread starter Jamin2112
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  • #1
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Homework Statement



Transform the given initial value problem into an initial value problem for the first two first order equations.

u'' + .25u' + 4u = 2cos(3t), u(0)=1, u'(0)=-2

Homework Equations



Nothing, really.

The Attempt at a Solution



x1=u , x2=u' => x2' = -.25x2 -4x1 + 2cos(3t); x1'=x2

There's the system. I don't understand the initial value part, though; and my professor didn't do any examples.

I know x1 and x2 are functions of t, so the second equation is saying that the derivative of x1 is x2, and x1'(0)=x2(0)=-2; x1(0)=1. Where do I go from here?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Homework Statement



Transform the given initial value problem into an initial value problem for the first two first order equations.

u'' + .25u' + 4u = 2cos(3t), u(0)=1, u'(0)=-2

Homework Equations



Nothing, really.

The Attempt at a Solution



x1=u , x2=u' => x2' = -.25x2 -4x1 + 2cos(3t); x1'=x2

There's the system. I don't understand the initial value part, though; and my professor didn't do any examples.

I know x1 and x2 are functions of t, so the second equation is saying that the derivative of x1 is x2, and x1'(0)=x2(0)=-2; x1(0)=1. Where do I go from here?

You don't go anywhere from there. The problem asked to change the u(t) equation into a coupled initial value problem for two first order equations. I think you did that with your x1(t) and x2(t).
 
  • #3
986
9
You don't go anywhere from there. The problem asked to change the u(t) equation into a coupled initial value problem for two first order equations. I think you did that with your x1(t) and x2(t).

I didn't read the problem! Hahaha!
 
  • #4
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6,648
I didn't read the problem! Hahaha!
If all else fails, read the instructions:smile:
 

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