Find the general solution of Bernoulli Eq

[Ready2GoXtr]

Problem:y' + y(1/x) = 3 x^2 y^2
Solution:p(x)= 1/x q(x) = 3x^2 <--- These are countinuous functions on the interval (0,+inf)

y^-2[y' + y(1/x) = 3 x^2 y^2] A

=> y'y^-2 + y^-1(1/x) = 3x^2

v = y^(-n+1) = y^-1 v' = -y^-2 y'

Plugin to problem A
v' + v/x = 3x^2 B

p(x) = 1/x q(x) =3x^2 <---- there's are continuous on the interval (0,+inf)

Find integral factor

h(x) = Int:[ 1/x dx] = ln(x) e^h(x) = e^(ln(x)) = x

multiply B through by factor

xv' + v = 3x^3
[xv]' = 3x^3
xv = Int:[3x^3 dx] = x^4 + C
solve for v

v = x^3 + C/x
replace v = y^-1

y^-1 = x^3 + C/x

y = (x^3 + C/x)^-1


Book listed solution = 2/(Cx - 3x^3)

Not sure what I am doing wrong there.

 

[Ready2GoXtr]

On a side question: does dy/dx = y' ?

 

[HallsofIvy]

That is standard notation. Surely you knew that?

A question back to you: "y(1/x)" means "y times 1/x" and not "y of 1/x"?