# Qualative look at diffy Q

## Homework Statement

The funciton with the given graph is a solution of one of the following differential equations. Decide which one is correct and justify your answer.

So does anyone know how I can draw this so you guys can see? Like a web site? The graph is non linear it is increasing up to what looks like a y intercept of 1 then it decreases non linearly in the first quadrant. It still increases for a tiny bit after is crosses the intercept.

## Homework Equations

A) y ' = 1+xy B) y' = -2xy C. y' = 1-2xy

## The Attempt at a Solution

I say it is C because that eq would have a y int of 1 which is what my graph looks like. and would have a neg slope in the first quadrant. But I don't know.

Anyone know a site so I can draw it so you can see?

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haruspex
Homework Helper
Gold Member
A) y ' = 1+xy B) y' = -2xy C. y' = 1-2xy
I say it is C because that eq would have a y int of 1
It would? Why?

HallsofIvy
Homework Helper

## Homework Statement

The funciton with the given graph is a solution of one of the following differential equations. Decide which one is correct and justify your answer.

So does anyone know how I can draw this so you guys can see? Like a web site? The graph is non linear it is increasing up to what looks like a y intercept of 1 then it decreases non linearly in the first quadrant. It still increases for a tiny bit after is crosses the intercept.

## Homework Equations

A) y ' = 1+xy B) y' = -2xy C. y' = 1-2xy

## The Attempt at a Solution

I say it is C because that eq would have a y int of 1 which is what my graph looks like. and would have a neg slope in the first quadrant. But I don't know.
I think you are misunderstanding what is given. The graph of y' would have a "y intercept of 1" but the graph you are given is of y, not y'. In (c), if x and y are both between 0 and 1/4 (so in the first quadrant) 1-2xy> 0 so the graph of y is increasing.

Anyone know a site so I can draw it so you can see?