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odie5533
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System of Differential Equations (Urgent)
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
[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.
Homework Statement
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
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.
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