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Using Jsxgraph to draw the vector field for a differential equation of the form $\frac{dy}{dx}=f(x,y)=x^2-y^2$, and also by moving the point $(x_0,y_0)$ you can see the solution curves going through that point.
If you want to modify $f(x,y)$ simply change the definition of $f(x,y)$ and that of the variable str in the user defined function testfield in Extensions and scripts. You have to use javascript notation for functions and powers in the definition of $f(x,y)$.
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Taxonomy: mathcentre
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Lois Rollings 7 years, 7 months ago
Created this as a copy of Solution curves and Vector fields 2 : dy/dx=x2−y2.| Name | Status | Author | Last Modified | |
|---|---|---|---|---|
| Solution curves and Vector fields 1 : dy/dx=sin(x-y) | Ready to use | Bill Foster | 10/04/2017 17:51 | |
| Solution curves and Vector fields 2 : dy/dx=x2−y2 | Ready to use | Bill Foster | 31/03/2017 21:24 | |
| Solution curves and Vector fields 3: dy/dx=x3−y3 | draft | Bill Foster | 31/03/2017 21:36 | |
| Solution curves and Vector fields 1 : dy/dx=1-sin(y) | draft | Bill Foster | 14/04/2017 11:51 | |
| Solution curves and Vector fields 1 : dy/dx=sin(x)-sin(y) | draft | Bill Foster | 10/04/2017 22:09 | |
| Lois's copy of Solution curves and Vector fields 2 : dy/dx=x2−y2 | draft | Lois Rollings | 14/09/2017 15:59 |
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Vector field for the differential equation \[\frac{dy}{dx}=f(x,y)=x^2-y^2\]
You change the function $f(x,y)$ by modifying the function $f(x,y)$ in the testfield function defined in Extensions and scripts.
You have to use Javascript notation for powers and standard functions when you define $f(x,y)$ in testfield.
If you want to see the solution curve through a particular point then drag the point labeled $(x_0,y_0)$ to it.
Use this tab to check that this question works as expected.
| Part | Test | Passed? |
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| Information only |