What is the solution for 2x^2y^'=piysin(pi/(2x)) ?

The solution for 2x^2y^'=piysin(pi/(2x)) is y=e^{cos(pi/(2x))+c_{1}}

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2x^2y^'=piysin(pi/(2x))

2 x^{2} y^{^{'}} = π y sin (\frac{π}{2 x})

y = e^{c o s (\frac{π}{2 x}) + c_{1}}

Solution steps

2 x^{2} y^{'} (x) = π y sin (\frac{π}{2 x})

Solve separable ODE:

y = e^{c o s (\frac{π}{2 x}) + c_{1}}

y = e^{c o s (\frac{π}{2 x}) + c_{1}}

x^{6 c o s (x)}

\frac{d}{d x} (8^{x^{4}})

\frac{x}{1 + x ^{2}}, \frac{x ^{2}}{1 + x ^{3}}, [0, 1]

\int \frac{8}{4 + 8 x} d x

\frac{180}{x ^{7}} - \frac{4}{x ^{3}}

What is the solution for 2x^2y^'=piysin(pi/(2x)) ?
The solution for 2x^2y^'=piysin(pi/(2x)) is y=e^{cos(pi/(2x))+c_{1}}