# Bernoulli Equation - Help?

CaptainJames

## Homework Statement

x^2y' + 2xy = 5y^3

## Homework Equations

Bernoulli Equation

## The Attempt at a Solution

so... v=y^-2, y=v^(-1/2) y' = -1/2v'v^(-3/2)

moving everything over...
y'/y^3 + 2/(xy^2) = 5/x^2

and plugging everything in...
-1/2v' + 2v/x = 5/x^2
v' - 4v/x = -10/x^2

v' - 4v/x = 0
dv/dx = 4v/x
dv/v = 4dx/x
ln v = 4 ln x + C
v = C(x)x^4
v' = C'(x)x^4 + C(x)4x^3

C'(x)x^4 + C(x)4x^3 - 4C(x)x^4/x = 5/x^2
C'(x)x^4 = 5/x^2
C'(x) = 5/x^6
C(x) = -x^-5 + C

v = (-x^-5 + C)x^4

y = 1/(v)^(1/2)

y^2 = 1/((-x^-5 + C)x^4)
... which is wrong. Could someone point me in the right direction?

Homework Helper

## Homework Statement

x^2y' + 2xy = 5y^3

## Homework Equations

Bernoulli Equation

## The Attempt at a Solution

so... v=y^-2, y=v^(-1/2) y' = -1/2v'v^(-3/2)

moving everything over...
y'/y^3 + 2/(xy^2) = 5/x^2

and plugging everything in...
-1/2v' + 2v/x = 5/x^2
v' - 4v/x = -10/x^2

v' - 4v/x = 0
dv/dx = 4v/x
dv/v = 4dx/x
ln v = 4 ln x + C
v = C(x)x^4

I think you are doing great up to the last line here. Why the x dependence in C?
I would just say v=C*x^4. This the general solution to the homogeneous equation v'-4v/x=0. Now you just want to add this to a particular solution (any one) of the inhomogeneous equation v' - 4v/x = -10/x^2. Can you find one?

CaptainJames
I think 2/x is a particular solution?

cause... you got, let's see...

-2/x^2 - 4*2/x^2 = -10/x^2

Okay so now what...??

Cx^4 + 2/x?

v=Cx^4 + 2/x

y = v^(-1/2)

y^2 = 1/(Cx^4+2/x)

or...

y^2 = x/(Cx^5 + 2) which is... CORRECT! thanks a bunch!

So all you have to do is find the general solution, then a particular solution? When is this applicable (only in cases where there is one derivative??)?

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