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## We need to reaarange the

We need to reaarange the differential equation so that it can be solved easily...

$$\left( x \csc\left( \frac{y}{x}\right) - y\right) dx + x dy = 0$$

$$x \csc\left( \frac{y}{x}\right)dx - ydx + xdy = 0$$

$$xdy = -x \csc\left( \frac{y}{x}\right)dx + ydx$$

Divide the terms of the differential equation by $x$, getting:

$$dy = - \csc\left( \frac{y}{x}\right)dx + \left( \frac{y}{x}\right)dx$$

Apparently this modified form of differential equation is a homogenous differential equation. So we let $v = \frac{y}{x}$, $y = vx$ and

$dy = vdx + xdv$. So we get:

$$dy = - \csc\left( \frac{y}{x}\right)dx + \left( \frac{y}{x}\right)dx$$

$$vdx + xdv = - \csc(v)dx + vdx$$

$$ xdv = - \csc(v)dx$$

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

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

The above differential equation is now a variable-separable type...so we get the integral of individual terms.

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

$$\int -sin(v)dv = \int \frac{1}{x} dx$$

$$cos(v) + c_1 = ln(x) + c_2$$

$$cos(v) = ln(x) + c_2 - c_1$$

$$cos(v) = ln(x) + C$$

Now getting the $v$:

$$cos(v) = ln(x) + C$$

$$v = cos^{-1}(ln(x) + C)$$

The solution of the given differential equation $\left( x \csc\left( \frac{y}{x}\right) - y\right) dx + x dy = 0$ would be:

$$y = vx$$

$$y = (cos^{-1}(ln(x) + C))x$$

$$y = xcos^{-1}(ln(x) + C)$$

Alternate ways of answering it are encouraged....