# I keep getting the same incorrect result in a differential equation and I don't know what I'm missing

So I've got the following differential equation, and this is how I'm trying to solve it.

$$x^2 + y^2 - 2xyy' = 0$$ $$y' = \frac{x^2 + y^2}{2xy}$$

Since it's a homogenous differential equation, I used $y=ux$, and got this: $$u' = \frac{1-u^2}{2ux}$$ $$\int \frac {2u\ du}{1-u^2} = \int \frac{dx}{x}$$

Using $1-u^2=t$ and $2u du = -dt$ I got the following:

$$\int-\frac{dt}{t} = \int\frac{dx}{x} \\ -\ln{t} = \ln{x} + c \\ 1-u^2 = x+e^c \\ y^2 + x^2 = x^2(x + e^c)$$

Yet, for some reason, the given solution is specifically this:

$$y^2 + x^2 = xc$$

Am I applying the method for solving these wrong or something? I'm lost. Thanks!

• If I'm not mistaken, you forgot $u$ in $y' = (u x)' = u' + u$ – Pavel Ievlev Dec 2 '17 at 15:51
• @PavelIevlev Isn't this correct?: $y' = (ux)' = u'x + ux' = xdu/dx + udx/dx = u'x + u$ – Robo Dec 2 '17 at 16:01
• Sorry. Everything's fine there. My mistake. – Pavel Ievlev Dec 2 '17 at 16:07

You got off to a great start! The following was all fine:

$$x^2 + y^2 - 2xyy' = 0$$ $$y' = \frac{x^2 + y^2}{2xy}$$ $$u' = \frac{1-u^2}{2ux}$$ $$\int \frac {2u\ du}{1-u^2} = \int \frac{dx}{x}$$ $$\int-\frac{dt}{t} = \int\frac{dx}{x}$$ $$-\ln{t} = \ln{x} + c$$

After that, though, you went off the rails. From there you should have:

$$-\ln(1-u^2) = \ln(x)+c$$ $$\ln\left(\frac{1}{1-u^2}\right)=\ln\left(e^cx\right)$$ $$\frac{1}{1-u^2}=e^cx$$ $$\frac{x^2}{x^2-u^2x^2}=e^cx$$ $$\frac{1}{x^2-y^2}=\frac{e^c}x$$ $$x^2-y^2=e^{-c}x$$ $$y^2=x^2-e^{-c}x$$ $$y^2=x^2+Cx$$

Another approach will be:

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

which is a linear equation with integration factor $e^{-\ln x}=1/x$. The solution is:

$$y^2 \frac{1}{x} = \int dx = x+C$$

The solution should be $$y^2 = x^2+ cx$$

The problem is here:

$-\ln (1 + u^2) = \ln x + \mathrm{const} \Rightarrow 1 + u^2 = \frac{C}{x},$

because $$-\ln x = \ln \frac{1}{x}.$$