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- Thread starter physicsss
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abercrombiems02

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first I'm going to determine how long I want to plot this for

>>t = linspace(0,10,1000);

this makes a vector t that is composed of 1000 equally spaced numbers between 0 and 10.

>>xt = your given function here;

>>xpt = take the derivative of the function above and code it here;

>>plot(xt,xpt) this plots xpt on the y-axis and xt on the x axis

>>xlabel('x(t)')

>>ylabel('x dot(t)')

>>title('single trajectory of x(t)')

>>grid on this will add a grid to your plot for easier reading

If you are not familiar with MATLAB I can explain how to input the functions if you're having trouble. However, if you need to do this in maple or whatever, just use the same approach, however I cannot help you with the syntax using that program as I have no experience using it.

The system above can also be plotted even easier using SIMULINK if you have any experience with that. In fact, when using simulink all you need to do is input the differential equation and your IC's and you can do just about anything to the output.

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physicsss

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Thanks, but I need to know how to do this by hand.

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saltydog

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I assume you mean:

[tex]x_1(t)=1/4e^{-t}+7/4e^{-2t}[/tex]

[tex]x_2(t)=-1/2e^{-t}+14/4e^{-2t}[/tex]

In this case, if the ODE was written as:

[tex]\frac{dx_1}{dt}=f(x_1,x_2)[/tex]

[tex]\frac{dx_2}{dt}=g(x_1,x_2)[/tex]

Then the slope field is drawn in terms of the vector field:

[tex]F[x,y]=\langle f(x,y),g(x,y) \rangle[/tex]

Then plot [itex]x_1(t)[/tex] and [itex]x_2(t)[/itex] parametrically in the same coordinate space. Just do 10 of them should be enough to get a rough plot say from t=0 to t=2. You know the points [itex](x_1(t),x_2(t))[/itex].

[tex]x_1(t)=1/4e^{-t}+7/4e^{-2t}[/tex]

[tex]x_2(t)=-1/2e^{-t}+14/4e^{-2t}[/tex]

In this case, if the ODE was written as:

[tex]\frac{dx_1}{dt}=f(x_1,x_2)[/tex]

[tex]\frac{dx_2}{dt}=g(x_1,x_2)[/tex]

Then the slope field is drawn in terms of the vector field:

[tex]F[x,y]=\langle f(x,y),g(x,y) \rangle[/tex]

Then plot [itex]x_1(t)[/tex] and [itex]x_2(t)[/itex] parametrically in the same coordinate space. Just do 10 of them should be enough to get a rough plot say from t=0 to t=2. You know the points [itex](x_1(t),x_2(t))[/itex].

Last edited:

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saltydog

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physicsss said:Does this look right?

Hey Physicsss. You know, I stated something incorrectly above regarding how to draw the slope field. I corrected it and hope that didn't cause confussion for you. Your plot looks Ok but it would be nice if you posted the ODE system. Also, it's a good idea to put an arrow at the end of the solution in order to identify the direction it's going.

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physicsss

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saltydog

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physicsss said:

Well, x1 and x2 are given in terms of t parametrically but we're graphing them as x2 in terms of x1. Calculate the derivatives of both then form the quotient:

[tex]\frac{\frac{dx_2}{dt}}{\frac{dx_1}{dt}}[/tex]

That becomes:

[tex]\frac{dx_2}{dx_1}[/tex]

That then is the slope of the graph we get. What do you suppose will be the slope you get?

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saltydog

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Physicsss, I made another mistake, Algebra this time. You know, the trajectory is not a straight line. It's just looks like one when you plot it from t=0 to t=2. Below is what it looks like closer to the origin (t>2). When we form the quotient of both derivatives, it's not constant as I initially thought but rather:

[tex]\frac{dy}{dx}=\frac{1/2e^{-t}-14/2e^{-2t}}{-1/4e^{-t}-7/2e^{-2t}}[/tex]

Which is:

[tex]\frac{2\left(1/4e^{-t}-7/2e^{-2t}\right)}{-\left(1/4e^{-t}+7/2e^{-2t}\right)}[/tex]

Sorry if I gave you the impression that it was a straight line.

[tex]\frac{dy}{dx}=\frac{1/2e^{-t}-14/2e^{-2t}}{-1/4e^{-t}-7/2e^{-2t}}[/tex]

Which is:

[tex]\frac{2\left(1/4e^{-t}-7/2e^{-2t}\right)}{-\left(1/4e^{-t}+7/2e^{-2t}\right)}[/tex]

Sorry if I gave you the impression that it was a straight line.

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