Solving $(\sqrt{xy}-x)\,dy+y\, dx=0$

Question

I am trying to solve the initial value problem $$ (\sqrt{xy}-x)\,dy+y\, dx=0 $$ $$ y(1)=0 $$ I have done the following: $y=xu$, $dy= u\,dx+x\, du$, $$ (\sqrt{x^2u}-x)(u\,dx+x\, du)+xu\, dx=0 $$ $$ (|x|\sqrt{u}-x)x\, du+|x|u\sqrt{u}\,dx=0 $$ ($|x|=x$ because the initial value $x=1$) $$ (\sqrt{u}-1)x^2\, du=-xu\sqrt{u}\,dx $$ $$ \frac{(\sqrt{u}-1)\, du}{u\sqrt{u}}=-\frac{dx}{x} $$ $$ \int\left( \frac1{u}-\frac1{u\sqrt{u}} \right)\, du=-\int\frac{dx}{x} $$ $$ \ln |u|+\frac2{\sqrt{u}}+\ln |x|= C $$ $$ \ln |ux|+\frac2{\sqrt{u}}= C $$ $$ \ln |y|+\frac2{\sqrt{y/x}}= C $$ I don't know how to obtain an initial value problem solution from this. I can't take a logarithm of zero and I can't divide by zero. Maybe I was wrong somewhere?

Upd. I was hinted to introduce a function $x(y)$ and solve the equation for it, but I obtained an equally useless result $$ \sqrt{x}=-\frac{y^2\sqrt{y}}4+C\sqrt{y} $$

Answer 1

Here: $$ \frac{(\sqrt{u}-1)\, du}{u\sqrt{u}}=-\frac{dx}{x} $$ takes place a division by zero. So we should check the solution $u=0$ or $y/x=0$, i.e. $y=0$. It is indeed a solution and it satisfies the initial conditions! So the answer is $y(x)=0$.

Answer 2

Mate,Your procedure and solution is totally right . But given initial condition is wrong ,since the solution curve $$\ln |y|+\frac2{\sqrt{y/x}}= C$$ Doesn't pass through point $(1,0)$ That's why are not getting any finite $C$ here..

Answer 3

The given ODE is homogeneous $$\frac{dy}{dx}=\frac{y}{x-\sqrt{xy}}.$$ Use $y=vx \implies y'=v+xv'$ We get $$\int \left(\frac{1}{v\sqrt{v}}-\frac{1}{v} \right)=\int \frac{dx}{x} \implies \frac{-2}{\sqrt{v}}=\ln v-\ln Cx \implies -2\sqrt{x}=\sqrt{y}\ln{Cy}$$