How did bernoulli think of his diff eqn sub? (1 Viewer)

[Average Boreduser]

bro my tutor just pulled let v=y^1-n out of thin air.... does anyone know how bernoulli dude came up with this sub?

 

[Average Boreduser]

trial and error maybe?

 

[liamkk112]

bro my tutor just pulled let v=y^1-n out of thin air.... does anyone know how bernoulli dude came up with this sub?

guessing ur talking about solving the eqn
y’+p(x)y = q(x)y^n
dividing by y^n:
y’y^-n +p(x)y^1-n = q(x)
now we can see that v = y^1-n is a good substitution because it will take place of the “coefficient” of p(x) and its derivative will be exactly y^-n which is attached to y’

 

[Drongoski]

The Bernoullis were a famous family of outstanding mathematians.

 

[Average Boreduser]

The Bernoullis were a famous family of outstanding mathematians.

guessing ur talking about solving the eqn
y’+p(x)y = q(x)y^n
dividing by y^n:
y’y^-n +p(x)y^1-n = q(x)
now we can see that v = y^1-n is a good substitution because it will take place of the “coefficient” of p(x) and its derivative will be exactly y^-n which is attached to y’

Is there like any solid proofs other than observable characteristics that may solidify the fact that the substituition works?

 

[liamkk112]

Is there like any solid proofs other than observable characteristics that may solidify the fact that the substituition works?

well it's just a nice substitution. its like with the integral of 1/(x^2+1), we know that subbing x as tan(theta) is a good substitution because it will lead to something that we can work with more easily. similarly we see that the substitution v = y^1-n works nicely as it happens to reduce the more difficult bernoulli equation to something more easily workable. it's just "intuition" same as any other substitution really