# Homogeneous equations- general solution

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shaaarmiii
Homogeneous equations- general solution

can u help me obtain the general equation of this
(( xcsc (y/x) - y ))dx + xdy = 0

fitzmerl duron 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....

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