# Thread: DE System won't check - can't find error

1. ## DE System won't check - can't find error

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 $u(t), v(t)$

$u'(t) + u(t) - v(t) = 0$ $u(0)=1$
$v'(t) - u(t) + v(t) = 2$ $v(0)=2$

My solution:

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

My work:

Laplace transform the system to get:
$U(s) = L(u(t))$, $V(s) = L(v(t))$

$sU(s) - 1 + U(s) - V(s) = 0$
$sV(s) - 2 - U(s) + V(s) = 2$

simplifying,
$(s+1)U(s) - V(s) = 1$
$- U(s) + (s+1)V(s) = 4$

solving for U(s), V(s)

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

partial fractions to inverse laplace later

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

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

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

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

2. Originally Posted by 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 $u(t), v(t)$

$u'(t) + u(t) - v(t) = 0$ $u(0)=1$
$v'(t) - u(t) + v(t) = 2$ [tex]v(0)=2[/math] (*)

My solution:

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

My work:

Laplace transform the system to get:
$U(s) = L(u(t))$, $V(s) = L(v(t))$

$sU(s) - 1 + U(s) - V(s) = 0$
$sV(s) - 2 - U(s) + V(s) = 2$ (**)

simplifying,
$(s+1)U(s) - V(s) = 1$
$- U(s) + (s+1)V(s) = 4$

solving for U(s), V(s)

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

partial fractions to inverse laplace later

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

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

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

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
My question is in (*) and (**). When taking the Laplace transform of the rhs of (*), why is $L\{2\} = 2$ and not $L\{2\} = \frac{2}{s}$

3. Originally Posted by Danny
My question is in (*) and (**). When taking the Laplace transform of the rhs of (*), why is $L\{2\} = 2$ and not $L\{2\} = \frac{2}{s}$
Because I am an idiot and didn't take the laplace transform of that constant!

ay yi yi...

Thanks very much for looking over this. That's my problem exactly!