## System of Differential Equations

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

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

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.
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 Recognitions: Gold Member Science Advisor Staff Emeritus Integrating both sides of x'+ y'= t2, you get x(t)+ y(t)= (1/3)t3. In particular, x(0)+ y(0)= 0.
 When doing that integration, isn't a constant introduced into the equation? $$x + y = \frac{t^3}{3} + C_{3}$$ $$x = \frac{t^3}{3} + C_{3} - y$$ $$x = \frac{t^3}{3} + C_{3} - 2t^2 + 5t - C_{1} - C_{2}E^{-t}$$ I also tried $$x + y = \frac{t^3}{3} + C_{3}$$ let t = 0 $$x(0) + y(0) = C_{3}$$ $$1 + 0 = C_{3}$$ $$x(t) = \frac{t^3}{3} + 1 - 2t^2 + 5t - C_{1} - C_{2}E^{-t}$$ x(0) = 1 $$1 = 1 - C_{1} - C_{2}$$ $$C_{1} = -C_{2}$$ Which is the same thing I got before. *EDIT* sorry, the initial condition for x was x(0) = 1.

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## System of Differential Equations

Oh, blast! You are right!
 So, is this solvable?

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