The solution to the Bernoulli problem This is what I have so far..... since y^-2

Question

The solution to the Bernoulli problem

This is what I have so far.....

since y^-2 is in the far side of the equation, the substitution will be u=y^3 or that y=cubic root(u).  From there I have a hard time getting the right side to cancel out and leave me with an expression free of y's or u's.  Any help is appreciated.

The solution to the Bernoulli problem y'+y = 4exy-2 f(x,y)-3e4x = C f(x,y) = This is what I have so far..... since y^-2 is in the far side of the equation, the substitution will be u=y^3 or that y=cubic root(u). From there I have a hard time getting the right side to cancel out and leave me with an expression free of y's or u's. Any help is appreciated.

Explanation / Answer

y^-2 -> n = -2

u = y^(1-n) = y^3 -> u' = 3y'y^2

y'+y = 4e^x y^-2 -> mutiply by y^2 :

y^2 y' + y^3 = 4e^x

1/3 u' + u = 4e^x

u' + 3u = 12e^x

multiply by e^(3x):

e^(3x)u' + 3e^(3x)u = 12e^(4x)

d/dx (e^(3x)u) = 12e^(4x)

e^(3x)u = int 12e^(4x) dx = 12/4 e^(4x) + C = 3e^(4x) + C

-> u = 3e^x + C.e^(-3x)

-> y^3 = 3e^x + C.e^(-3x)

-> (y^3 - 3e^x)/e^(-3x) = C

e^(3x)y^3 - 3e^(4x) = C

so f(x,y) = e^(3x)y^3

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