Systems of Linear Differential Equations using Laplace
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HallsofIvy
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The Laplace transform of function f is \(\displaystyle F(s)= \int_0^\infty f(t)e^{-st}dx\). In particular, The Laplace transform of y' is sY(s)- y(0) where Y(s) is the Laplace transform of y. The Laplace transform of f(t)= t is \(\displaystyle F(s)= \frac{1}{s^2}\).
So applying the Laplace transform to both sides of both equations,
\(\displaystyle sX- 1+ Y= \frac{1}{s^2}\) and
\(\displaystyle sY- 1+ 4X= 0\)
Treat those as two linear equations in X and Y:
\(\displaystyle sX+ Y= 1+ \frac{1}{s^2}\)
\(\displaystyle 4X+sY= 1\)
Multiply the first equation by s and subtract the second to eliminate Y:
\(\displaystyle (s^2- 4)X= s+ \frac{1}{s^2}- 1\)
\(\displaystyle X= \frac{s+ \frac{1}{s^2}- 1}{s^2- 4}= \frac{s^3- s^2+ 1}{s^2(s^2- 4)}\)
Now use "partial fractions" to reduce the right side to a sum of terms you can look up in a table of Laplace transforms.
So applying the Laplace transform to both sides of both equations,
\(\displaystyle sX- 1+ Y= \frac{1}{s^2}\) and
\(\displaystyle sY- 1+ 4X= 0\)
Treat those as two linear equations in X and Y:
\(\displaystyle sX+ Y= 1+ \frac{1}{s^2}\)
\(\displaystyle 4X+sY= 1\)
Multiply the first equation by s and subtract the second to eliminate Y:
\(\displaystyle (s^2- 4)X= s+ \frac{1}{s^2}- 1\)
\(\displaystyle X= \frac{s+ \frac{1}{s^2}- 1}{s^2- 4}= \frac{s^3- s^2+ 1}{s^2(s^2- 4)}\)
Now use "partial fractions" to reduce the right side to a sum of terms you can look up in a table of Laplace transforms.
Hello HallsofIvy,
Thank you for the reply! I just want to clarify, If I multiply s to the first equation, shouldn't the 1/s^2 be 1/s?
Then using partial fraction, I got the result,
-1/4s + 7/8(s+2) + 3/8(s-2),
I'm assuming I have to do Inverse Laplace to get the value of x(t), but how can I get the value in terms of y(t)?
Thank you for the help. Really appreciate it
Thank you for the reply! I just want to clarify, If I multiply s to the first equation, shouldn't the 1/s^2 be 1/s?
Then using partial fraction, I got the result,
-1/4s + 7/8(s+2) + 3/8(s-2),
I'm assuming I have to do Inverse Laplace to get the value of x(t), but how can I get the value in terms of y(t)?
Thank you for the help. Really appreciate it
HallsofIvy
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- Jan 27, 2012
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Yes, it should. Sorry about that.
Of course, with a very simple example like this it is simpler to solve the equations without using the Laplace transform.
If you differentiate the first equation, x'+ y= t again, you have x''+ y'= 1. From the second equation, y'+ 4x= 0, y'= -4x so that equation is x''- 4x= 1. Solve that, then solve for y= t- x'. You can use that to check your answer.
Of course, with a very simple example like this it is simpler to solve the equations without using the Laplace transform.
If you differentiate the first equation, x'+ y= t again, you have x''+ y'= 1. From the second equation, y'+ 4x= 0, y'= -4x so that equation is x''- 4x= 1. Solve that, then solve for y= t- x'. You can use that to check your answer.
I checked the answer using your suggestion and I got almost the same answer from when I used the Laplace Transform (one of the sign is in positive, and my answer on the Laplace is negative), I didn't bother solving for the constants though when I checked so that might be it.
Thanks a lot! I learned so much
Thanks a lot! I learned so much