Question

which of the following could be the slope field for the differential equation dy/dx=x^2y

Answer 1

THESE ARE THE OPTIONS https://scontent-atl3-1.xx.fbcdn.net/hphotos-xat1/v/t34.0-12/13096107_1332124626803842_6555411910540396088_n.jpg?oh=840b0991021551c5d07df1bfc045fa21&oe=5721463D

Answer 2

You can solve the differential equation since is separable: https://en.wikipedia.org/wiki/Examples_of_differential_equations#Separable_first-order_ordinary_differential_equations

Answer 3

I still don't get it... @Bobo-i-bo

Answer 4

@The_Real_Rapper

Answer 5

graph iv

Answer 6

\[\frac{ dy }{ dx } = x^2y\]
So by seperating the variables, using bad maths we can say:
\[\frac{ 1 }{ y }dy=x^2dx\]
So:
\[\int\limits \frac 1 y dy = \int\limits x^2 dx\]
And so if you solve the integrals on both sides and rearrange, you can get y as a function of x

Answer 7

When you were in algebra you learned \(y=mx+b\) and m was the slope. Rise over run, right? \[m=\frac{\Delta y}{\Delta x}\]

similarly here we have
\[\frac{dy}{dx} = x^2y\]
so you can imagine this representing the slope at every point in the xy plane.
at the point \((-1, 2)\) you can plug this in to get:

\[\frac{dy}{dx} = (-1)^22 = 2\] so you find the point on the grid (-1,2) and write a tiny mark with the same slope as the line with m=2 would have.

See how right at that point I made a little line where it looks like it has m=2? Remember, rise over run, so \(m=\frac{2}{1}\) means rise by 2 run to the right by 1. You just want this slope only at this point.

Repeat this process for a handful of points and you should get one of the pictures!

Answer 8

So would it be graph III? @Kainui

Answer 9

yes i understood it wrong yes graph III