Find the general solution for the differential equation $xy^{'}\cos {\frac{y}{x}}=y\cos {\frac{y}{x}}-x$ Find the general solution for the differential equation $xy^{'}\cos {\frac{y}{x}}=y\cos {\frac{y}{x}}-x$
After substitution $t=\frac{y}{x},y^{'}=t^{'}x+t$ the equation is 
$$x(t^{'}x+t)\cos t=x(t\cos t-1)\Rightarrow (t^{'}x+t)\cos t=(t\cos t-1)$$
How to separate variables by $dt$ and $dx$?
 A: Notice, one can separate variables as follows  $$(t'x+t)\cos t=(t\cos t-1)$$
$$t'x\cos t+t\cos t=t\cos t-1$$
$$t'x\cos t=-1$$
$$t'\cos t=-\frac{1}{x}$$
$$\cos t\frac{dt}{dx}=-\frac{1}{x}$$
$$\cos t\ dt=-\frac{dx}{x}$$
I hope you can take it from here.
A: $$xy'(x)\cos\left(\frac{y(x)}{x}\right)=y(x)\cos\left(\frac{y(x)}{x}\right)-x\Longleftrightarrow$$
$$x\cdot\frac{\text{d}y(x)}{\text{d}x}\cdot\cos\left(\frac{y(x)}{x}\right)=y(x)\cos\left(\frac{y(x)}{x}\right)-x\Longleftrightarrow$$

Let $y(x)=xv(x)$, which gives $\frac{\text{d}y(x)}{\text{d}x}=v(x)+\frac{\text{d}v(x)}{\text{d}x}$:

$$x\cos(v(x))\left(x\cdot\frac{\text{d}v(x)}{\text{d}x}+v(x)\right)=-x+x\cos(v(x))v(x)\Longleftrightarrow$$
$$x\cos(v(x))\left(x\cdot\frac{\text{d}v(x)}{\text{d}x}+v(x)\right)=x(\cos(v(x))v(x)-1)\Longleftrightarrow$$
$$\frac{\text{d}v(x)}{\text{d}x}=-\frac{\sec(v(x))}{x}\Longleftrightarrow$$
$$\cos(v(x))\cdot\frac{\text{d}v(x)}{\text{d}x}=-\frac{1}{x}\Longleftrightarrow$$
$$\int\cos(v(x))\cdot\frac{\text{d}v(x)}{\text{d}x}\space\text{d}x=-\int\frac{1}{x}\space\text{d}x\Longleftrightarrow$$
$$\sin(v(x))=-\ln\left|x\right|+\text{C}\Longleftrightarrow$$
$$v(x)=\arcsin\left(-\ln\left|x\right|+\text{C}\right)\Longleftrightarrow$$
$$y(x)=x\arcsin\left(-\ln\left|x\right|+\text{C}\right)\Longleftrightarrow$$
$$y(x)=-x\arcsin\left(\ln\left|x\right|-\text{C}\right)$$
A: write the Eq in the form 
$$
y'=\frac{y}{x}-\sec(\frac{y}{x})
$$
then, using $y=xt$ we have $y'=t+x\frac{dt}{dx}$. Hence
$$
t+x\frac{dt}{dx}=t-\sec(t)\rightarrow \cos(t)dt=-\frac{dx}{x}\rightarrow \sin(t)=-\ln|x|+C
$$
and the general solution is 
$$
\sin(y/x)=-\ln|x|+c.
$$
A: Hint: write that equation in the form $y'\cos(y/x)=y/x\cos(y/x)-1$ for $x\ne 0$ and by setting $y=ux$ we get
$$(u'x+u)\cos(u)=u\cos(u)-1$$
