Solve System of Diff Eqns Mistake: Find Answer

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In summary, the solution to the given system of differential equations is y = x = e^t. The mistake in the original solution was due to incorrect Laplace inversions. Laplace transforms can be defined for complex numbers, as shown by the alternative solution for y = e^t using complex numbers.
  • #1
frggr
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I'm not sure where I am making the mistake was hoping you guys could help.
-------------------------------------------------------------------------------------------------------
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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  • #2
Y = (s - 5) / [(s-3)² - 4] (OK)
Better simplify !
(s-3)²-4 = (s-5)(s-1)
Y=1/(s-1)
 
  • #3
welcome to pf!

hi frggr! welcome to pf! :smile:
frggr said:
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' :wink:
 
  • #4
thank you both for the reply, and yea I was just trying to get practice with Laplace transforms.

JJacquelin said:
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?
 
  • #5
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.
 
  • #6
JJacquelin said:
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):
[tex]
I = \int_{0}^{\infty } e^{-st}f(t)dt
[/tex][tex]
I = \int_{0}^{\infty } e^{-st} / e^{3t}sin(2t)dt
[/tex][tex]
I = \int_{0}^{\infty } e^{(3-s)t}sin(2t) dt
[/tex][tex]
I = \int _{0}^{\infty }e^{-(s-3)t}sin(2t)
[/tex]


Calculating the integral:
[tex]
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
[/tex]
[tex]
u = e^{-s^*t} \ \ du =-s^*e^{-s^*t}dt
[/tex]
[tex]
dv = sin(2t)dt\ \ v = -\frac{1}{2} cos(2t)
[/tex]

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

[tex]
\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
[/tex]
[tex]
\therefore \frac{s^*^2 - 4}{4}I = [0 + \frac{1}{2}] -[0 - 0]
[/tex]
[tex]
I =\frac{4}{s^*^2 - 4} * \frac{1}{2} = \frac{2}{s^*^2 - 4} = \frac{2}{(s-3)^2 - 4}
[/tex]
 
Last edited:
  • #7
Laplace Transform of exp(3t)sin(2t) isn't 2/((s-3)-4)
it's 2/((s-3)²+4)
 
  • #8
JJacquelin said:
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?
 
  • #9
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)
 

What is a system of differential equations?

A system of differential equations is a set of equations that describe how a set of variables change over time in relation to each other. These equations often involve derivatives and are used to model many different phenomena in fields such as physics, engineering, and biology.

What does it mean to solve a system of differential equations?

Solving a system of differential equations means finding a set of functions that satisfy all of the equations in the system. This can involve finding a general solution or a particular solution that satisfies certain initial conditions.

What is a common mistake when solving a system of differential equations?

A common mistake when solving a system of differential equations is to ignore or forget to account for initial conditions. These conditions are important because they determine the specific solutions that satisfy the equations in the system.

How can I check if my solution to a system of differential equations is correct?

To check if your solution to a system of differential equations is correct, you can substitute the functions into the original equations and see if they satisfy all of the equations. You can also use numerical methods to approximate the solutions and compare them to your solution.

What are some methods for solving systems of differential equations?

Some common methods for solving systems of differential equations include separation of variables, substitution, and using matrix methods. Other techniques such as Laplace transforms and numerical methods may also be used depending on the complexity of the system.

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