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(A) The given differential equation is $\frac{dy}{dx} - \frac{y}{x} + \csc\left(\frac{y}{x}\right) = 0$.This can be rewritten as $\frac{dy}{dx} = \fra

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$\cos\left(\frac{y}{x}\right) = \log|ex|$

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(A) The given differential equation is $\frac{dy}{dx} - \frac{y}{x} + \csc\left(\frac{y}{x}\right) = 0$.This can be rewritten as $\frac{dy}{dx} = \frac{y}{x} - \csc\left(\frac{y}{x}\right)$.Since this is a homogeneous differential equation,we substitute $y = vx$,which implies $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting these into the equation,we get $v + x\frac{dv}{dx} = v - \csc(v)$.Simplifying,we have $x\frac{dv}{dx} = -\csc(v)$,which leads to $-\sin(v) dv = \frac{dx}{x}$.Integrating both sides,we get $\int -\sin(v) dv = \int \frac{1}{x} dx$.This results in $\cos(v) = \log|x| + C$.Substituting $v = \frac{y}{x}$,we get $\cos\left(\frac{y}{x}\right) = \log|x| + C$.Given $y = 0$ when $x = 1$,we have $\cos(0) = \log|1| + C$,which gives $1 = 0 + C$,so $C = 1$.Thus,$\cos\left(\frac{y}{x}\right) = \log|x| + 1 = \log|x| + \log(e) = \log|ex|$.Therefore,the particular solution is $\cos\left(\frac{y}{x}\right) = \log|ex|$.

Find the particular solution satisfying the given condition:
$\frac{dy}{dx} - \frac{y}{x} + \csc\left(\frac{y}{x}\right) = 0$; $y = 0$ when $x = 1$.

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Find the particular solution satisfying the given condition:$\frac{dy}{dx} - \frac{y}{x} + \csc\left(\frac{y}{x}\right) = 0$; $y = 0$ when $x = 1$.