Bernoulli Differential Equation (with a missing solution)

[@holyshit922]

Bernoulli equation has separable integrating factor
We can use integrating factor directly to Bernoulli equation and then integrating factor will be in the separable form

[@moonsmoonsloverg1rl]

How do you do that?

[@holyshit922]

@@moonsmoonsloverg1rl in separable form means that you look for integrating factor in the form
mu(x,y) = phi(x)psi(y)
then plug in into condition for exact equation
Suppose we have following equation
P(x,y)dx+Q(x,y)dy = 0
Let multiply it by undetermined function mu(x,y)
mu(x,y)P(x,y)dx+mu(x,y)Q(x,y)dy = 0
Condition for exact equation is
partial mu(x,y)P(x,y)/partial y = partial mu(x,y)Q(x,y)/partial x
partial mu(x,y)P(x,y)/partial y - partial mu(x,y)Q(x,y)/partial x
Now if you assume that mu(x,y) = phi(x)psi(y)
Then after calculations assume that
d phi(x)/phi(x) = f(x)dx
d psi(y)/psi(y) = g(y)dy
You have following equatiion for that type of integrating factor
partial P(x,y)/partial y -partial Q(x,y)/partial x = Q(x,y)f(x) - P(x,y)g(y)
In Bernoulli equation you have
(p(x)y - q(x)y^r)dx + dy = 0
1*f(x) - (p(x)y - q(x)y^r)g(y) = 0
Now if you let
f(x) = Ap(x)
g(y) = B/y
You will have following equation
p(x) - rq(x)y^{r-1} = Ap(x) - B(p(x) - q(x)y^{r-1})
p(x) - rq(x)y^{r-1} = (A - B)p(x) + Bq(x)y^{r-1}
And you have system of equations for A and B
A - B = 1
B = -r
I skipped some calculations because it is the comment

[@zzwag]

Wow thank you so much for this!!

[@blackpenredpen]

zZwag do u think this format is okay? Or u prefer me to write the steps along the way?

[@Blackfir333]

blackpenredpen It's easier for me to understand when you write the steps as you explain :)

[@zzwag]

Hi! Yeah as a student, I find it easier to see the problem without any steps already written out. This way I can take a second to think about the problem before seeing the solution. However either way works fine since you explain concepts really well.

[@blackpenredpen]

Thank you both for your inputs. As you can see, even I had the steps on the board already and it still took me 9+ min for this vid. I just wanted to try this format to shorten the vid

[@dennisivanchavez]

blackpenredpenbluepen

[@tahaabujrad7806]

What if we let the constant equal to inf. ,by doing that we get 1/(f(x)+inf) which is zero .

[@pranavsuren9489]

Constant shouldn't be infinite
It never makes sense as you would have infinite as a term. The constant is just some finite real number

[@JesusGarcia-ox3jj]

do some reduction order differential equations.

[@blackpenredpen]

Here's one
th-cam.com/video/qw7lsWSkfGA/w-d-xo.html

[@FarhanAli-dq8eh]

How does (C/e^(-2x)) turn into Ce^(2x)?

[@hamsterdam1942]

since 1/(a^b)=a^(-b)

[@marcushendriksen8415]

1/e^(2x) = e^(-2x), therefore, 1/e^(-2x) = e^(2x)

[@rhversity5965]

Wow 0 dislikes first video ive seen that

[@enochakintoke8634]

What about the music😢?

[@pompei2]

Why you do not prove, that missing solutions no any more?

[@mateusarguelhodacunha8425]

nice!

[@andreguimaraes9347]

Wouldn't y=0 be part of the family of solutions, since when c->Infinity, y->0???

[@hamsterdam1942]

no, since inf isnt a num