A system of 1st order diffy q's

[Jamin2112]

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?

 

[Dick]

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).

 

[Jamin2112]

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!

 

[Mark44]

I didn't read the problem! Hahaha!

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