Solve System of Diff Eqns Mistake: Find Answer

[frggr]

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 ee^3t
y = e^3tcos(-2t) - e^3tsin(-2t)

y = e^3tcos(2t) + e^3tsin(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 = e^3tcos(2t) + e^3tsin(2t)

Answer SHOULD BE
y = x = e^t
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Which does make sense because when plugged into originial eqn e^t works, where as my answer does not.

 

[JJacquelin]

Y = (s - 5) / [(s-3)² - 4] (OK)
Better simplify !
(s-3)²-4 = (s-5)(s-1)
Y=1/(s-1)

 

[tiny-tim]

welcome to pf!

hi frggr! welcome to pf!

x' = 3x - 2y; x₀=1
y' = 3y - 2x; y₀=1

i assume you wanted to practise Laplace Transforms, but if you just want a quickie solution, just looking at those equations immediately suggests that the obvious thing to do is to write out x' + y' and x' - y'

 

[frggr]

thank you both for the reply, and yea I was just trying to get practice with Laplace transforms.

Y = (s - 5) / [(s-3)² - 4] (OK)
Better simplify
(s-3)²-4 = (s-5)(s-1)
Y=1/(s-1)

Heh, guess I shouldn't have missed that. Question now is why did my solution come out wrong? -- Simplifying shouldn't really change the outcome just how difficult it is to get there right?

 

[JJacquelin]

Question now is why did my solution come out wrong? -- Simplifying shouldn't really change the outcome just how difficult it is to get there right?

There are some mistakes in your Laplace inversions.
The inverse Laplace of 2/((s-3)²-4) is exp(3t)sinh(2t)
The inverse Laplace of (s-3)/((s-3)²-4) is exp(3t)cosh(2t)
Then exp(3t)cosh(2t)-exp(3t)sinh(2t) = exp(3t)exp(-2t) = exp(t)
So, the outcome is the same.

 

[frggr]

There are some mistakes in your Laplace inversions.
The inverse Laplace of 2/((s-3)²-4) is exp(3t)sinh(2t)
The inverse Laplace of (s-3)/((s-3)²-4) is exp(3t)cosh(2t)
Then exp(3t)cosh(2t)-exp(3t)sinh(2t) = exp(3t)exp(-2t) = exp(t)
So, the outcome is the same.

I'm more lost now than when I started. -> I must be doing Laplace transforms completely wrong. Or am using the wrong definition to calculate them because :

I tried computing exp(3t)sin(2t) by definition and got 2/((s-3)^2-4), Any explanations would be appreciated

I've supplied my calculations below

Laplace Transform of exp(3t)sin(2t):
<br /> I = \int_{0}^{\infty } e^{-st}f(t)dt<br /><br /> I = \int_{0}^{\infty } e^{-st} / e^{3t}sin(2t)dt<br /><br /> I = \int_{0}^{\infty } e^{(3-s)t}sin(2t) dt<br /><br /> I = \int _{0}^{\infty }e^{-(s-3)t}sin(2t)<br />


Calculating the integral:
<br /> I = \int_{0}^{\infty } e^{-s^*t}sin(-2t)dt \therefore -I = \int_{0}^{\infty } e^{-s^*t}sin(2t)dt,\ \ where\ \ s^* = s-3<br />
<br /> u = e^{-s^*t} \ \ du =-s^*e^{-s^*t}dt<br />
<br /> dv = sin(2t)dt\ \ v = -\frac{1}{2} cos(2t)<br />

<br /> \therefore -I =-e^{-s^*t}\frac{1}{2} cos(2t)]_{0}^{\infty} + \frac{-s^*}{2} \int_{0}^{\infty}cos(2t)e^{-s^*t}<br />
<br /> u = e^{-s^*t}\ \ du=-s^*e^{-s^*t}dt <br />
<br /> dv = cos(2t) \ \ v = \frac{1}{2}sin(2t)<br />

<br /> \therefore -I = [-\frac{1}{2}e^{-s^*t} cos(2t)]_{0}^{\infty} - [\frac{-s^*}{4}e^{-s^*t}sin(2t)]_{0}^{\infty} +\frac{s^*^2}{4} I<br />
<br /> \therefore \frac{s^*^2 - 4}{4}I = [0 + \frac{1}{2}] -[0 - 0] <br />
<br /> I =\frac{4}{s^*^2 - 4} * \frac{1}{2} = \frac{2}{s^*^2 - 4} = \frac{2}{(s-3)^2 - 4}<br />

 

[JJacquelin]

Laplace Transform of exp(3t)sin(2t) isn't 2/((s-3)-4)
it's 2/((s-3)²+4)

 

[frggr]

Laplace Transform of exp(3t)sin(2t) isn't 2/((s-3)-4)
it's 2/((s-3)²+4)

Ahh, thanks.

Not sure why I thought (-2)^2 would be -4. My bad.
Do have one last question though I don;t really care if it's answered.

Would

e^(3t) cos(2i) + i e(3t)sin(2i) be a valid solution, or are Laplace transforms only defined for Real Numbers?

 

[JJacquelin]

e^(3t) cos(2i) + i e(3t)sin(2i) be a valid solution, or are Laplace transforms only defined for Real Numbers?

I suppose that your question is about
y = e^(3t) cos(2it) + i e^(3t)sin(2it)
If so, y = e^(3t)*(cos(2it)+i*sin(2it))
y = e^(3t)*e^(i*2it) = e^(3t)*e^(-2t)
y = e^t which is a valid solution
Another way :
cos(2it) = cosh(2t)
sin(2it) = i sinh(2t)