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The given differential equation is: $x^{2}\frac{dy}{dx}-xy=x^{2}cos^{2}(\frac{y}{2x})$ Divide throughout by $x^2$: $\frac{dy}{dx} - \frac{y}{x} = cos^{2}(\frac{y}{2x})$
This is a homogeneous differential equation. Let $y = vx$, so $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substitute $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$ into the equation: $v + x\frac{dv}{dx} - v = cos^{2}(\frac{vx}{2x})$ $x\frac{dv}{dx} = cos^{2}(\frac{v}{2})$
Separate the variables: $\frac{dv}{cos^{2}(\frac{v}{2})} = \frac{dx}{x}$ Integrate both sides: $\int \frac{dv}{cos^{2}(\frac{v}{2})} = \int \frac{dx}{x}$ $\int sec^{2}(\frac{v}{2}) dv = \int \frac{dx}{x}$ $2tan(\frac{v}{2}) = ln|x| + C$
$2tan(\frac{y}{2x}) = ln|x| + C$
Given that when $x=1$, $y=\frac{\pi}{2}$: $2tan(\frac{\pi/2}{2(1)}) = ln|1| + C$ $2tan(\frac{\pi}{4}) = 0 + C$ $2(1) = C$ $C = 2$
Substitute $C = 2$ into the general solution: $2tan(\frac{y}{2x}) = ln|x| + 2$ $tan(\frac{y}{2x}) = \frac{1}{2}ln|x| + 1$ $\frac{y}{2x} = tan^{-1}(\frac{1}{2}ln|x| + 1)$ $y = 2x \cdot tan^{-1}(\frac{1}{2}ln|x| + 1)$
Final Answer: $y = 2x \cdot tan^{-1}(\frac{1}{2}ln|x| + 1)$
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