What does it mean when a questions asks me to sketch the graph of dy/dx against y, and then sketch the graph of dy/dx against x? How do I do this?

Question

I was given a homework question that asks for me to sketch the graph of dy/dx against x and sketch the graph of dy/dx against y (we were only given dy/dx= 3y^2+9y). I don't understand what it means to sketch dy/dx against x and dy/dx against y, or how I would do it. Are they different? I am especially confused about dy/dx against x, since there is no x in the equation.

Answer 1

graphing g(y)=3y²+9y is not really a problem (dy/dx=g(y))

now know also from equation and graph that when y=1,,,,,y'(x)=12 for example

g(y)=dy/dx=y'(x)

dx=dy/(3y(y+3)

dx=(1/3)[1/(y(y+3)]dy=(1/3)[((1/3)/y)-((1/3)/(y+3))]dy

x=(1/9)ln|y|-(1/9)ln|y+3|+C

x=(1/9)ln|y/(y+3)|+C,,, integrating using partial fractions

must use absolute value for ln function since -3<y<0 values are valid in definition of dy/dx.

letting C=0,,,this can be plotted and at y=1 it is plausible for dy/dx to be 12.

y/(y+3)=ke^9x

trying to find dy/dx will have y terms and y=f(x) not known

so plotting dy/dx with x as independent variable

would have to be done point by point and initial condition (the C term) will affect the values.

Answer 2

This may not be the only solution (and I hope other tutors will chime in):

You can solve the differential equation for y in terms of x (use partial fractions).

Then differentiate to get y' and plot it.

I think the solution set has 3 branches.

The derivative is 0 when y=0 and when y=-3.

The solution y'=0 is a source and when y'=-3 it is a sink.

Between these 2 solutions the solution is shaped like a backwards S

These branches can be APPROXIMATED by

y=e^-2x, y=3-e^2x and y=3/(1-e^-2x

The branches outside -3<y<0 actually each have an inflection point which is not shown by the approximate solutions given above.

8/18/2023

It may be of no use to the original questioner, but maybe to others who review this question in the future:

The solution to this problem is

y=3/(Ke^-9x-1).