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

#### 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 $$\displaystyle u(t), v(t)$$

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

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
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}$$\)

• kpizle

#### kpizle

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!