DE System won't check - can't find error

Oct 2008
43
4
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 \(\displaystyle u(t), v(t)\)

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

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
 

Jester

MHF Helper
Dec 2008
2,470
1,255
Conway AR
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 \(\displaystyle u(t), v(t)\)

\(\displaystyle u'(t) + u(t) - v(t) = 0\) \(\displaystyle u(0)=1\)
\(\displaystyle v'(t) - u(t) + v(t) = 2\) \(\displaystyle 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\)
\(\displaystyle
My question is in (*) and (**). When taking the Laplace transform of the rhs of (*), why is \(\displaystyle L\{2\} = 2\) and not \(\displaystyle L\{2\} = \frac{2}{s}\)\)
 
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Reactions: kpizle
Oct 2008
43
4
My question is in (*) and (**). When taking the Laplace transform of the rhs of (*), why is \(\displaystyle L\{2\} = 2\) and not \(\displaystyle L\{2\} = \frac{2}{s}\)
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!