What is the solution for y^'= y/x+tan(y/x) ?

The solution for y^'= y/x+tan(y/x) is y=arcsin(c_{1}x)x+2pinx,y=pix+arcsin(-c_{1}x)x+2pinx

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y^'= y/x+tan(y/x)

y^{^{'}} = \frac{y}{x} + tan (\frac{y}{x})

y = arcsin (c_{1} x) x + 2 πn x, y = π x + arcsin (- c_{1} x) x + 2 πn x

Solution steps

y^{'} (x) = \frac{y}{x} + tan (\frac{y}{x})

Solve using substitution:

y = arcsin (c_{1} x) x + 2 πn x, y = π x + arcsin (- c_{1} x) x + 2 πn x

y = arcsin (c_{1} x) x + 2 πn x, y = π x + arcsin (- c_{1} x) x + 2 πn x

4 y^{^{''}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 3

\frac{\partial}{\partial x} (x e^{- a x^{2}})

\frac{d}{d x} (a (2 e^{- 3 x} x - 3 e^{- 3 x} x^{2}))

\frac{\partial}{\partial x} (- sin (4 x))

f (x) = x^{2} + 2 x + 3

What is the solution for y^'= y/x+tan(y/x) ?
The solution for y^'= y/x+tan(y/x) is y=arcsin(c_{1}x)x+2pinx,y=pix+arcsin(-c_{1}x)x+2pinx