Solving Bernoulli's Differential Equation

[Dusty912]

Homework Statement

xy(dx)=(y²+x)dy

Homework Equations

integrating factor : u(x)=e^∫p(x)dx
standard form of linear DE: dy/dx + P(x)y=Q(x)
standard form of bernoulli's differential equation: dy/dx + P(x)y=Q(x)yⁿ
change of variables v=y^1-n

The Attempt at a Solution

xy(dy)=(y²+x)dx
xy(dy/dx)=y² +x
dy/dx=y/x +1/y
dy/dx-y/x=y^-1

yⁿ=y^-1
y^-n=y¹

multipliying both sides by y¹ yeilds:
y¹(dy/dx) -y²/x=1

using change of variable v=y^1-n=y²
dv/dx=(2y)(dy/dx)
(1/2)(dv/dx)=(y)(dy/dx)
and subbing the results yeilds:
(1/2)(dv/dx)-v/x=1

using an integrating factor to solve the resulting linear differential equation
u(x)=e^∫p(x)dx
u(x)=e^-∫(1/x)dx=1/x

multiplying both sides of the equation by the integrating factor yeilds:
(1/x)[(1/2)(dv/dx)-v/x]=1/x
(d/dx)[v/x]=1/x

integrating both sides:
(1/2)∫(v/x)d=∫1/xdx
(1/2)v/x=ln|x| + C
v=2ln|x| +C

now subbing back in y^2]
y^2]=2ln|x| + C

Unfortunately the answer is :

y²=-2x+Cx²

where did I go wrong?
Thanks ahead of time, you guys rock

 

[LCKurtz]

Homework Statement

xy(dx)=(y²+x)dy

Homework Equations

integrating factor : u(x)=e^∫p(x)dx
standard form of linear DE: dy/dx + P(x)y=Q(x)
standard form of bernoulli's differential equation: dy/dx + P(x)y=Q(x)yⁿ
change of variables v=y^1-n

The Attempt at a Solution

xy(dy)=(y²+x)dx
xy(dy/dx)=y² +x
dy/dx=y/x +1/y
dy/dx-y/x=y^-1

yⁿ=y^-1
y^-n=y¹

multipliying both sides by y¹ yeilds:
y¹(dy/dx) -y²/x=1

using change of variable v=y^1-n=y²
dv/dx=(2y)(dy/dx)
(1/2)(dv/dx)=(y)(dy/dx)
and subbing the results yeilds:
(1/2)(dv/dx)-v/x=1

using an integrating factor to solve the resulting linear differential equation

Your equation is not in the proper form for the integrating factor. The leading coefficient must be $1$, so you need to multiply your equation by $2$ before continuing.

 

[Dusty912]

thanks a bunch, I got it now