The Bernoulli equation is almost identical to the standard linear ODE.

Compare to the basic linear ODE,

Meanwhile, the Riccati equation takes the form

which in special cases is of Bernoulli type if

, and linear if

. But in general each type takes a different method to solve. From now on, I'll abbreviate the coefficient functions as

for brevity.
For Bernoulli equations, the standard approach is to write


and substitute

. This makes

, so the ODE is rewritten as

and the equation is now linear in

.
The Riccati equation, on the other hand, requires a different substitution. Set

, so that

. Then you have



Next, setting

, so that

, allows you to write this as a linear second-order equation. You have



where

and

.
If you just need the slope intercept form it is y=mx+b
Answer:
install socratic it Will give you all answer no kiding
Answer:
the slope is 2/3x
Step-by-step explanation:
how i got the answer is using rise over run.
Answer:


Step-by-step explanation:

First, let's put this equation into slope-intercept form by subtracting
from both sides of the equation:

Let's graph this (see the attached photo.)
As you see we have a straight line. If you are able to graph a straight line, the domain and range will be all real numbers. This can be represented as:

- or -

~

- or -