Differential Equations: Direction Field

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SUMMARY

The discussion focuses on sketching a direction field for the differential equation $y' = x - y + 1$. Participants emphasize the importance of understanding the mapping of the sign of $y'$ in the x-y plane and suggest computing slopes at integer lattice points for clarity. It is recommended to initially draw the direction field by hand before using a calculator or software for visualization. The equation can be analyzed by setting $y' = m$, leading to the line $y = x + 1 - m$ which helps determine the behavior of the solution curves.

PREREQUISITES
  • Understanding of differential equations, specifically first-order linear equations.
  • Familiarity with direction fields and their significance in visualizing solutions.
  • Basic graphing skills, particularly with Cartesian coordinates.
  • Experience with graphing calculators or software capable of plotting direction fields.
NEXT STEPS
  • Learn how to compute direction fields for various differential equations using software like Desmos or GeoGebra.
  • Explore the concept of stability in solution curves of differential equations.
  • Study the method of integrating factors for solving first-order linear differential equations.
  • Investigate the use of numerical methods, such as Euler's method, for approximating solutions to differential equations.
USEFUL FOR

Students studying differential equations, educators teaching calculus concepts, and anyone interested in visualizing mathematical solutions through direction fields.

ineedhelpnow
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Sketch a direction field for the differential equation. Then use it to sketch three solution curves.
$y'=x-y+1$

I really need help drawing this, I'm super confused. :confused:
 
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ineedhelpnow said:
Sketch a direction field for the differential equation. Then use it to sketch three solution curves.
$y'=x-y+1$

I really need help drawing this, I'm super confused. :confused:

The so called 'direction field' is simply the mapping of the sign of y' in the x-y plane. So setting y' = m = const, You obtain...

y' = m -> y = x + 1 - m (1)

The (1) means that in all the points where y > x + 1 is y' < 0, in all the points where y = x + 1 is y' = 0 and in all the points where y < x + 1 is y' > 0...

Kind regards

$\chi$ $\sigma$
 
Last edited:
what my book does is plug in values for x and y into the equation and then they compute a whole bunch of different slopes. Should I make my graph from like $-2\le x \le 2$ and $-2\le y \le 2$ ?
 
is there a way to do it on my calculator?
 
ineedhelpnow said:
what my book does is plug in values for x and y into the equation and then they compute a whole bunch of different slopes. Should I make my graph from like $-2\le x \le 2$ and $-2\le y \le 2$ ?

That's what I would do...and for simplicity only compute the slope at lattice points, that is, those points whose coordinates are integers. This will give you 25 points at which to compute a slope.

I imagine your calculator can draw direction fields...back when I was a student, we had to program our calculators to do this. :D
 
oh i think i got it
 
i had it but i mistakenly closed the document and i keep putting in the equation but it won't show up anymore. I am putting in the equation of the graph but it's also asking me for initial conditions.
 
Last edited:
I really recommend you do this problem by hand first, and only then look at a computer generated plot of the field. You get much more of a feel for what's going on by actually getting in there and doing it yourself. :D
 

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