Unable to solve the following question

Question

I am trying to solve the following question but I can not figure out a method to solve it. The question is:

$$x+\sqrt{(y^2) - (xy) \frac{dy}{dx}} = y, y(1/2) = 1$$

Can somebody help me and guide me through? If you could solve it I would get the idea and solve the remaining questions.

Answer 1

For $x\gt0,$ $x+\sqrt{y(x)^2-xy(x)y'(x)}=y(x)$ is equivalent to $$x+\sqrt{x^2\left[\left(\frac{y(x)}{x}\right)^2-\frac{y(x)}{x}y'(x)\right]}=y(x)=x+x\sqrt{\left[\left(\frac{y(x)}{x}\right)^2-\frac{y(x)}{x}y'(x)\right]},$$ which is equivalent to $$\frac{y(x)}{x}=1+\sqrt{\left[\left(\frac{y(x)}{x}\right)^2-\frac{y(x)}{x}y'(x)\right]}.$$ Let $z(x)=y(x)/x,$ such that $z(1/2)=2,$ and $y'(x)=xz'(x)+z(x)$ $$z(x)=1+\sqrt{z(x)^2-xz(x)z'(x)-z(x)^2}=1+\sqrt{-xz(x)z'(x)}.$$ Since $x\gt0,$ we have that $z(x)z'(x)\lt0,$ but since $\sqrt{-xz(x)z'(x)}\geq0,$ we have that $1+\sqrt{-xz(x)z'(x)}\geq1,$ so $z(x)\geq1.$ Therefore, it is the case that $z'(x)\lt0.$ With this in mind, the differential equation implies $$[z(x)-1]^2=-xz(x)z'(x),$$ which is equivalent to $$-\frac1{x}=\frac{z(x)}{[z(x)-1]^2}z'(x)=\left(\frac1{z(x)-1}+\frac1{[z(x)-1]^2}\right)z'(x)=\left[\ln(z-1)-\frac1{z-1}\right]'(x).$$ Given the condition that $x\gt0$ and $z(x)\geq1,$ we have that $$\ln[z(x)-1]-\frac1{z(x)-1}=-\ln(x)+C,$$ which is equivalent to $$[z(x)-1]e^{-\frac1{z(x)-1}}=\frac{e^C}{x}.$$ Since $z(1/2)=2,$ we have that $2e^C=e^{-1},$ so $e^C=\frac1{2e}.$ So we have that $$\frac1{2ex}=[z(x)-1]e^{-\frac1{z(x)-1}},$$ which is equivalent to $$\frac1{z(x)-1}e^{\frac1{z(x)-1}}=2ex$$ given the restrictions. Here, we can use the Lambert W function, obtaining us that $$\frac1{z(x)-1}=W(2ex),$$ which is equivalent to $$z(x)=1+\frac1{W(2ex)},$$ which is equivalent to $$y(x)=x\left[1+\frac1{W(2ex)}\right].$$

Answer 2

After squaring:

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

$x-2y=-y\frac{dy}{dx}$

$(x-2y)dx+ydy=0$

Let $y=\lambda x$ since every term is of linear degree. Then $dy=xd\lambda+\lambda dx$

$(1-2\lambda)dx+\lambda (xd\lambda+\lambda dx)=(\lambda -1)^2dx+\lambda xd\lambda=0$

$\frac{dx}{x}=\frac{(-\lambda+1 - 1) d\lambda}{(\lambda -1)^2}$

$\ln{x}=-\ln{|\lambda -1|}+\frac{1}{\lambda -1} +c_1$

$x= \frac{c_2e^{1/(\lambda -1)}}{\lambda -1}=\frac{c_2xe^{\frac{x}{y-x}}}{y-x}$

So cancelling $x$:

$y-x=c_2xe^{\frac{x}{y-x}}$