System of Differential Equations

odie5533

1. The problem statement, all variables and given/known data
Solve for y(t). You need not find x(t).
[tex]2x' + y' - y = t[/tex]
[tex]x' + y' = t^2[/tex]
x(0) = 1, y(0) = 0

3. The attempt at a solution
[tex]2Dx + (D - 1)[y] = t[/tex]
[tex]Dx + Dy = t^2[/tex]
[tex]2D^2x + (D^2 - D)[y] = 1[/tex]
[tex]2D^2x + 2D^2y = 4t[/tex]
[tex](D^2 + D)[y] = 4t - 1[/tex]
[tex]y'' + y' = 4t - 1[/tex]
I solved the above equation using undetermined coefficients. It's a lot of writing, so I'll just put the answer here:
[tex]y(t) = 2t^2 - 5t + C_{1} + C_{2}E^{-t}[/tex]
Using the initial value y(0) = 0
[tex]0 = C_{1} + C_{2}[/tex]
[tex]C_{1} = -C_{2}[/tex]

I'm stuck at applying the other initial value. It says you don't need to find x(t), so I was wondering if there is a way to do this without going through and solving for x(t) then solving for the constants and such.

HallsofIvy

Integrating both sides of x'+ y'= t², you get x(t)+ y(t)= (1/3)t³. In particular, x(0)+ y(0)= 0.

odie5533

When doing that integration, isn't a constant introduced into the equation?
[tex]x + y = \frac{t^3}{3} + C_{3}[/tex]
[tex]x = \frac{t^3}{3} + C_{3} - y[/tex]
[tex]x = \frac{t^3}{3} + C_{3} - 2t^2 + 5t - C_{1} - C_{2}E^{-t}[/tex]


I also tried
[tex] x + y = \frac{t^3}{3} + C_{3}[/tex]
let t = 0
[tex]x(0) + y(0) = C_{3}[/tex]
[tex]1 + 0 = C_{3}[/tex]
[tex]x(t) = \frac{t^3}{3} + 1 - 2t^2 + 5t - C_{1} - C_{2}E^{-t}[/tex]
x(0) = 1
[tex]1 = 1 - C_{1} - C_{2}[/tex]
[tex]C_{1} = -C_{2}[/tex]
Which is the same thing I got before.

*EDIT* sorry, the initial condition for x was x(0) = 1.

HallsofIvy

Oh, blast! You are right!

odie5533

So, is this solvable?

Dick

If x(t) solves your equations, then so does x(t)+C. A value for x(0) doesn't constrain the solution in any way. Are you sure the boundary condition isn't for x'(0)? BTW you could avoid introducing an extra derivative and just written a first order equation for y.

HallsofIvy

Oh, very good, Dick. I'm embarassed I didn't see that. (Don't worry, in few minutes I will have convinced myself that I did!)

From the second equation, x'= t²- y^2 so you can eliminate x' from the first equation.