- #1
frggr
- 7
- 0
I'm not sure where I am making the mistake was hoping you guys could help.
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Q:
x' = 3x - 2y; x0=1
y' = 3y - 2x; y0=1
My Sol:
1) Take Laplace Transform of both sides.
X:=L{x}, Y := L{y}
Xs - x0 = 3X - 2Y
Ys - y0 = 3Y - 2X
Xs - 1 = 3X - 2Y
Ys - 1 = 3Y - 2X
Xs - 3X = 1 - 2Y
Ys - 3Y = 1 - 2X
(s-3)X + 2Y = 1 Eqn 1
(s-3)Y + 2X = 1 Eqn 2
2Eqn 1 - (s-3)Eqn 2
2(s-3)X + 4Y - (s-3)2Y - 2(s-3)X = 2 - (s-3)
[4 - (s-3)2]Y = 2 - s + 3
Y = -(s - 5) / [4 - (s-3)2]
Y = (s - 5) / [(s-3)2 - 4]
Y = (s - 3) / [(s-3)2 - 4] - 2 / [(s-3)2 - 4]
L-1{Y} = L-1{(s - 3) / [(s-3)2 - 4]} - L-1{2 / [(s-3)2 - 4]}
These are sin and cos shifted by 3, so Sin and cos times ee3t
y = e3tcos(-2t) - e3tsin(-2t)
y = e3tcos(2t) + e3tsin(2t)
(s-3)X + 2Y = 1 Eqn 1
(s-3)Y + 2X = 1 Eqn 2
(s-3)Y + 2X = (s-3)X + 2Y
(s-5)Y = (s-5)X
Y = X :. y=x
y = x = e3tcos(2t) + e3tsin(2t)
Answer SHOULD BE
y = x = et
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Which does make sense because when plugged into originial eqn et works, where as my answer does not.
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Q:
x' = 3x - 2y; x0=1
y' = 3y - 2x; y0=1
My Sol:
1) Take Laplace Transform of both sides.
X:=L{x}, Y := L{y}
Xs - x0 = 3X - 2Y
Ys - y0 = 3Y - 2X
Xs - 1 = 3X - 2Y
Ys - 1 = 3Y - 2X
Xs - 3X = 1 - 2Y
Ys - 3Y = 1 - 2X
(s-3)X + 2Y = 1 Eqn 1
(s-3)Y + 2X = 1 Eqn 2
2Eqn 1 - (s-3)Eqn 2
2(s-3)X + 4Y - (s-3)2Y - 2(s-3)X = 2 - (s-3)
[4 - (s-3)2]Y = 2 - s + 3
Y = -(s - 5) / [4 - (s-3)2]
Y = (s - 5) / [(s-3)2 - 4]
Y = (s - 3) / [(s-3)2 - 4] - 2 / [(s-3)2 - 4]
L-1{Y} = L-1{(s - 3) / [(s-3)2 - 4]} - L-1{2 / [(s-3)2 - 4]}
These are sin and cos shifted by 3, so Sin and cos times ee3t
y = e3tcos(-2t) - e3tsin(-2t)
y = e3tcos(2t) + e3tsin(2t)
(s-3)X + 2Y = 1 Eqn 1
(s-3)Y + 2X = 1 Eqn 2
(s-3)Y + 2X = (s-3)X + 2Y
(s-5)Y = (s-5)X
Y = X :. y=x
y = x = e3tcos(2t) + e3tsin(2t)
Answer SHOULD BE
y = x = et
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Which does make sense because when plugged into originial eqn et works, where as my answer does not.