Bernoulli equations getting lost just before they become linear odes

[MickGrif]

I have two examples of bernoulli odes;

(a) dy/dx +xy = 6xy²

I have this solved down as far as dz/dx +xy/2y^1/2 = 3x, now I know the next line is dz/dx + x²/2 = 3x but I have know idea as to what happens between these two lines.

Likewise in
(b) dy/dx = e^xy = y²e^x I have solved down as far as dz/dx + e^x/y = e^x with the next line reading dz/dx = e^x (z?) = -e^x but again I am unsure of how this was reached (I do realize that you can solve this particular equation using variable separable method but I would like to know how to do it with Bernoulli as well


I know how to finish off the equation using linear odes after, it is just this point at which I am confused could anyone help me out?

 

[Deleted member 4993]

I have two examples of bernoulli odes;

(a) dy/dx +xy = 6xy²

I have this solved down as far as dz/dx +xy/2y^1/2 = 3x,

now I know the next line is dz/dx + x²/2 = 3x

How do you know that? What is 'z'?

but I have know idea as to what happens between these two lines.(your definition of z will guide you about the in between steps)

Likewise in
(b) dy/dx = e^xy = y²e^x (Something wrong here)I have solved down as far as

dz/dx + e^x/y = e^x with the next line reading dz/dx = e^x (z?) = -e^x

Again define your 'z'

but again I am unsure of how this was reached (I do realize that you can solve this particular equation using variable separable method but I would like to know how to do it with Bernoulli as well


I know how to finish off the equation using linear odes after, it is just this point at which I am confused could anyone help me out?

Also try to write your equations with grouping symbols - following PEMDAS. Otherwise, you'll write something - while trying mean something else (e.g. xy/2y^1/2 means x * y/2 * y^1/2 -I don't think you meant that)

 

[galactus]

I will step through a Bernoulli for part a.

A Bernoulli DE has the form \(\displaystyle \frac{dy}{dx}+P(x)y=f(x)y^{n}\)

We use the sub \(\displaystyle u=y^{1-n}\) as long as \(\displaystyle n\neq 0, \;\ n\neq 1\)

So, since we have n=2, we use \(\displaystyle y=u^{-1}\)

\(\displaystyle \frac{dy}{dx}=-u^{-2}\frac{du}{dx}\), by the chain rule.

Subbing into our given DE we get:

\(\displaystyle -u^{-2}\frac{du}{dx}+xu^{-1}=6xu^{-2}\)

\(\displaystyle \frac{du}{dx}-xu=-6x\)

The integrating factor is \(\displaystyle e^{-\int xdx}=e^{-\frac{x^{2}}{2}}\)

\(\displaystyle \frac{d}{dx}[ue^{-\frac{x^{2}}{2}}]=-6xe^{-\frac{x^{2}}{2}}\)

Integrate:

\(\displaystyle ue^{-\frac{x^{2}}{2}}=6e^{-\frac{x^{2}}{2}}+C\)

\(\displaystyle u=6+Ce^{\frac{x^{2}}{2}}\)

Backsubbing our \(\displaystyle u=\frac{1}{y}\), we get:

\(\displaystyle \displaystyle \boxed{y=\frac{1}{6+Ce^{\frac{x^{2}}{2}}}}\)