Laplace transform for system DFE

[wxrebecca]

dx/dt= -x+y
dy/dt= 2x
x(0)=0
y(0)=1

I'm not familiar how to solve a system like this. Somebody please help?

Thanks

 

[mjpam]

Is it an equation of one or several independent variables?

 

[wxrebecca]

I think it's several independent variables. However, the question didn't say so

 

[mjpam]

I think it's several independent variables. However, the question didn't say so

Which of the following did it look like:

\frac{dx}{dt}=-x+y

\frac{dy}{dt}=2x

x(0)=0

y(0)=1

or

\frac{\partial x}{\partial t}=-x+y

\frac{\partial y}{\partial t}=2x

x(0)=0

y(0)=1

 

[wxrebecca]

Which of the following did it look like:

\frac{dx}{dt}=-x+y

\frac{dy}{dt}=2x

x(0)=0

y(0)=1

or

\frac{\partial x}{\partial t}=-x+y

\frac{\partial y}{\partial t}=2x

x(0)=0

y(0)=1

it's the first one.
but what's the differences between these two exactly?

 

[HallsofIvy]

I'm not sure why mjpam asked the question in the first place. The only differentiation is with respect to t so even if there were other (unstated) independent variables, it would make no difference to the solution. If solutions to differential equations (and other problems) depended on variables we knew nothing about, we would never be able to solve any problem!

wxrebecca, you titled this "Laplace transform for system DFE". If you really are required to use Laplace transform, then, you just take the Laplace transform of each equation, L(dx/dt)= -L(x)+ L(y) and L(dy/dt)= 2L(x). That will reduce to two algebraic equations for L(x) and L(y). Once you have solved for those, use a transform table to find x and y as functions of t.

Personally, I have never liked Laplace transform. What I would do is differentiate the first equation again:
\frac{d^2x}{dt^2}= -\frac{dx}{dt}+ \frac{dy}{dt}
From the second equation, dy/dt= 2x so that equation is
\frac{d^2x}{dt^2}= -\frac{dx}{dt}+ 2x
or
\frac{d^2x}{dt^2}+ \frac{dx}{dt}- 2x= 0
a second order linear equation with constant coefficients. Once you have solved that, you can use the fact that
y= \frac{dx}{dt}+ x
from the first equation to solve for y.

 

[mjpam]

I'm not sure why mjpam asked the question in the first place. The only differentiation is with respect to t so even if there were other (unstated) independent variables, it would make no difference to the solution. If solutions to differential equations (and other problems) depended on variables we knew nothing about, we would never be able to solve any problem!

If y is a function of x,

\frac{dy}{dx}=\frac{d^{2}x}{dt^2}

It seems to simplify the equation.

 

[trambolin]

Write it as this
<br /> \begin{pmatrix}<br /> \dot x(t)\\ \dot y(t)<br /> \end{pmatrix}=<br /> \begin{bmatrix}<br /> -1 &amp;1\\2 &amp;0<br /> \end{bmatrix}<br /> \begin{pmatrix}<br /> x(t) \\y(t)<br /> \end{pmatrix}<br />

Just like in the scalar case the solution of this differential equation is given as
<br /> \begin{pmatrix}<br /> x(t)\\y(t)<br /> \end{pmatrix} = e^{At}<br /> \begin{pmatrix}<br /> x(0)\\y(0)<br /> \end{pmatrix}<br />
So you have to find the matrix exponential and you are done. It is not element by element exponantiation by the way. I'll leave that to you.

 

[maverick280857]

dx/dt= -x+y
dy/dt= 2x
x(0)=0
y(0)=1

I'm not familiar how to solve a system like this. Somebody please help?

Thanks

HallsofIvy has solved your problem. You just need to take the Laplace transform of both equations, and use the expressions for the laplce transform of the derivative of a function:

\mathcal{L}(dy/dt) = sY(s) - y(0)

This is a system of coupled equations. You can either decouple them, and get one equation (in x or y) which is of second order, or you can solve them by taking the Laplace transform of each, and expressing X(s) in terms of Y(s) or vice versa.

If you decouple them to get a second order equation (in say x) you must know x(0) and x'(0). You can get x'(0) from the first differential equation by plugging in the known values of x(0) and y(0). You already know x(0). There is no need to do a matrix exponential.