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**kpizle** Hi. I finished solving a system of DEs using Laplace transform some time ago. The values I figured do not solve the system when plugged back into the original. However, I cannot find my error. I am sure it is something small, but I have been laboring for over an hour and cannot find where I went wrong. I feel like I've double-checked everything, but that cannot be so because the answers do not check.

The problem:

Solve the following for $\displaystyle u(t), v(t)$

$\displaystyle u'(t) + u(t) - v(t) = 0$ $\displaystyle u(0)=1$

$\displaystyle v'(t) - u(t) + v(t) = 2$ [tex]v(0)=2[/math] (*)

My solution:

$\displaystyle u(t) = \frac{5 - 3e^{-2t}}{2}$

$\displaystyle v(t) = \frac{3e^{-2t} + 5}{2}$

My work:

Laplace transform the system to get:

$\displaystyle U(s) = L(u(t))$, $\displaystyle V(s) = L(v(t))$

$\displaystyle sU(s) - 1 + U(s) - V(s) = 0$

$\displaystyle sV(s) - 2 - U(s) + V(s) = 2$ (**)

simplifying,

$\displaystyle (s+1)U(s) - V(s) = 1$

$\displaystyle - U(s) + (s+1)V(s) = 4$

solving for U(s), V(s)

$\displaystyle U(s) = \frac{s+5}{s(s+2)}$

$\displaystyle V(s) = \frac{4s+5}{s(s+2)}$

partial fractions to inverse laplace later

$\displaystyle U(s) = \frac{5}{(2s)} - \frac{3}{2(s+2)}$

$\displaystyle V(s) = \frac{3}{2(s+2)} + \frac{5}{(2s)}$

inverse laplace, obtaining u(t), v(t)

$\displaystyle u(t) = (5/2) - (3/2)e^{-2t}$

$\displaystyle v(t) = (3/2)e^{-2t} + (5/2)$

simplify and you get what I wrote above.

The problem is, when I plug them into the system, both equations work out to 0...the second one is not equal to 2.

I have poured over this and cannot figure it out.

Can someone spot my error?

Thanks