The problem:

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

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

\(\displaystyle v'(t) - u(t) + v(t) = 2\) \(\displaystyle v(0)=2\)

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