Nonlinear First Order ODE: Bernoulli Equation with n = 2

[Mangoes]

Homework Statement

(y^2 + xy)dx - x^2dy = 0

The Attempt at a Solution

Put it into derivative form.

y^2 + xy - x^2 \frac{dy}{dx} = 0

\frac{dy}{dx} - \frac{y^2}{x^2} - \frac{xy}{x^2} = 0

\frac{dy}{dx} + \frac{-1}{x}y = \frac{1}{x^2}y^2

I recognized this as a Bernoulli equation where n = 2.

\frac{dy}{dx}\frac{1}{y^2} + \frac{-1}{x}\frac{1}{y} = \frac{1}{x^2}

Plan to make a substitution of v = y^{-1} and \frac{dv}{dx} = -y^{-2} \frac{dy}{dx}

-\frac{dv}{dx} + \frac{-1}{x}v = \frac{1}{x^2}

\frac{dv}{dx} + \frac{1}{x}v = \frac{-1}{x^2}

u(x) = e^{\int{\frac{1}{x}}dx}

u(x) = x

\int{x\frac{dv}{dx} + v} = \int{\frac{-1}{x}}

LHS is product of product rule.

xv = -lnx + c

Since v = y^(-1)

\frac{x}{y} = ln{\frac{c}{x}}

You could arrange the above equation into various forms, but I see no way to arrange it into the form in the answer key:

x + yln{|x|} = cy

I've checked again and again and can't see where I'm going wrong with this...

 

[voko]

Recall what the logarithm of a ratio is.

 

[HallsofIvy]

Homework Statement

(y^2 + xy)dx - x^2dy = 0

The Attempt at a Solution

Put it into derivative form.

y^2 + xy - x^2 \frac{dy}{dx} = 0

\frac{dy}{dx} - \frac{y^2}{x^2} - \frac{xy}{x^2} = 0

\frac{dy}{dx} + \frac{-1}{x}y = \frac{1}{x^2}y^2

I recognized this as a Bernoulli equation where n = 2.

\frac{dy}{dx}\frac{1}{y^2} + \frac{-1}{x}\frac{1}{y} = \frac{1}{x^2}

Plan to make a substitution of v = y^{-1} and \frac{dv}{dx} = -y^{-2} \frac{dy}{dx}

-\frac{dv}{dx} + \frac{-1}{x}v = \frac{1}{x^2}

\frac{dv}{dx} + \frac{1}{x}v = \frac{-1}{x^2}

u(x) = e^{\int{\frac{1}{x}}dx}

u(x) = x

\int{x\frac{dv}{dx} + v} = \int{\frac{-1}{x}}

LHS is product of product rule.

xv = -lnx + c

Since v = y^(-1)

\frac{x}{y} = ln{\frac{c}{x}}

Strictly speaking this should be
\frac{x}{y} = ln{\left|\frac{c}{x}\right|}
and, of course, ln|c/x|= ln|c|- ln|x|

You could arrange the above equation into various forms, but I see no way to arrange it into the form in the answer key:

x + yln{|x|} = cy

I've checked again and again and can't see where I'm going wrong with this...