How To Draw Direction Fields For Differential Equations

TheChaz

I would:
1. Solve for y' (simply by subtracting 3y)
2. Pick points (t, y) in the t-y airplane, and calculate the correct-hand-side of the equation.
3. So for/at each point (t, y), brand a little line segment with slope = y'

For instance, at the betoken (0, 0), y ' = -3(0) + (0) + e^(-2*0) = 1
So at the point (0, 0), I would depict a fiddling line segment (non longer than, say ane-1.5 units) with slope ~1.
At the point (0, y), the gradient is -3y + 1, so that can help yous with stride iii, at to the lowest degree for the line t = 0

TheChaz

Thanks for the reply.

1. I considered doing all that, but I tin can only stay on the y-centrality for so long. I attempted to exercise (1, 2) and ended up with y' = two(2) + 3e^1 which is probably somewhere just nether 28.
2. Is it supposed to exist that high? That just seems pretty weird.

3. And are yous expected to just keep inputting points into the equation to determine the graph?

1. I simply emphasized the y-axis for its simplicity. There aren't shortcuts for other lines (maybe the line y = t...)
two. 4 + e is not 28! Information technology's around 6.7, but honestly, information technology'due south difficult to tell the difference betwixt a teeny line with gradient half dozen and a teeny line with slope 28 anyway...
3. yep.

Prove It

How on earth are you supposed to graph direction fields for these super complicated differential equations?

I kind of understood bones ones, simply now we're learning equations like:

y' + 3y = t + east^(-2t)

How would you describe the direction field for that DE? (past manus)

Sorry I'm only learning this in school and I'm very lost here. Thank you for any help.

Why not just solve the DE? It's showtime social club linear...

\(\displaystyle \displaystyle \frac{dy}{dt} + 3y = t + due east^{-2t}\)

The integrating factor is \(\displaystyle \displaystyle e^{\int{3\,dt}} = e^{3t}\), so multiplying both sides past this gives

\(\displaystyle \displaystyle \begin{align*} eastward^{3t}\frac{dy}{dt} + 3e^{3t}y &= t\,eastward^{3t} + due east^t \\ \frac{d}{dt}\left(eastward^{3t}y\right) &= t\,eastward^{3t} + east^t \\ e^{3t}y &= \int{t\,due east^{3t} + due east^t\,dt} \\ eastward^{3t}y &= \frac{i}{three}t\,eastward^{3t} - \int{\frac{1}{3}e^{3t}\,dt} + e^t + C \\ eastward^{3t}y &= \frac{1}{3}t\,e^{3t} - \frac{ane}{nine}e^{3t} + e^t + C \\ y &= \frac{1}{3}t - \frac{1}{ix} + e^{-2t} + Ce^{-3t} \terminate{align*} \)