# Homogeneous DE Problem

1. Homework Statement

I'm told that this is homogenous
$$(x^2-xy)y'+y^2 = 0$$

2. The attempt at a solution

This is going to be very painful for me to type out...

$$(x^2-xy)\frac{dy}{dx} = -y^2$$

$$\frac{dy}{dx} = \frac{-y^2}{(x^2-xy)}$$

$$\frac{dy}{dx} = \frac{-y^2}{(x^2-xy)}$$

$$v+x\frac{dv}{dx} = \frac{-x^2v^2}{(x^2-x^2v)}$$

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

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

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

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

$$\int-v^-^2dv = \int\frac {dx}{x}$$

$$v^-^1 =ln|x|+C$$

$$\frac{y^-^1}{x^-^1} = ln|x|+C$$

$$y^-^1 = x ln|x|+C$$

$$\frac{1}{y} = x ln|x|+C$$

$$\frac{1}{y} = \frac {ln|x|+C}{x}$$

$$0 = \frac {y ln|x|+C}{x}$$

$$-Cx = y ln|x|$$

$$xln|y| - y = Cx$$

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cristo
Staff Emeritus
1. Homework Statement

I'm told that this is homogenous
$$(x^2-xy)y'+y^2 = 0$$

2. The attempt at a solution

This is going to be very painful for me to type out...

$$(x^2-xy)\frac{dy}{dx} = -y^2$$

$$\frac{dy}{dx} = \frac{-y^2}{(x^2-xy)}$$

$$\frac{dy}{dx} = \frac{-y^2}{(x^2-xy)}$$

$$v+x\frac{dv}{dx} = \frac{-x^2v^2}{(x^2-x^2v)}$$

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

$$v+x\frac{dv}{dx} = \frac{-v^2}{(1-v)}$$
Try going from here again. Add another line in if you need to. Put all the v's over a common denominator before you multiply by dx.

HallsofIvy
Homework Helper
1. Homework Statement

I'm told that this is homogenous
$$(x^2-xy)y'+y^2 = 0$$

2. The attempt at a solution

This is going to be very painful for me to type out...

$$(x^2-xy)\frac{dy}{dx} = -y^2$$

$$\frac{dy}{dx} = \frac{-y^2}{(x^2-xy)}$$

$$\frac{dy}{dx} = \frac{-y^2}{(x^2-xy)}$$

$$v+x\frac{dv}{dx} = \frac{-x^2v^2}{(x^2-x^2v)}$$

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

$$v+x\frac{dv}{dx} = \frac{-v^2}{(1-v)}$$
$$\frac{-v^2}{1-v}-v= \frac{-v^2-v+v^2}{1-v}= \frac{-v}{1-v}$$
The left hand side in your next line is wrong.

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

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

$$\int-v^-^2dv = \int\frac {dx}{x}$$

$$v^-^1 =ln|x|+C$$

$$\frac{y^-^1}{x^-^1} = ln|x|+C$$

$$y^-^1 = x ln|x|+C$$

$$\frac{1}{y} = x ln|x|+C$$

$$\frac{1}{y} = \frac {ln|x|+C}{x}$$

$$0 = \frac {y ln|x|+C}{x}$$

$$-Cx = y ln|x|$$

$$xln|y| - y = Cx$$

Oh I understand the error, thanks, I'll work it out from there

I've messed up again...

$$v+x\frac{dv}{dx} = \frac{-v^2}{(1-v)}$$
$$\frac{v(1-v)}{-v^2}dv = \frac{dx}{(x)}$$
$$\frac{v-v^2}{-v^2}dv = \frac{dx}{(x)}$$
$$\int v^-^1-1 dv = \int \frac{dx}{(x)}$$
$$ln|v|-x = ln|x|+C$$
$$ln|\frac {y}{x}|-x = ln|x|+C$$

and no need to go any farter.. I know its wrong...

is it cus this line

$$\frac{v(1-v)}{-v^2}dv = \frac{dx}{(x)}$$

should look like this?

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

cristo
Staff Emeritus
$$v+x\frac{dv}{dx} = \frac{-v^2}{(1-v)}$$
$$x\frac{dv}{dx}=\frac{-v^2}{(1-v)}-v=\frac{-v^2-v(1-v)}{(1-v)}$$