What is the solution for y^'+5yx= x/2-xy ?

The solution for y^'+5yx= x/2-xy is y=(e^{-3x^2+c_{1}})/(12)+1/12

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y^'+5yx= x/2-xy

y^{^{'}} + 5 y x = \frac{x}{2} - x y

y = \frac{e ^{- 3 x^{2} + c_{1}}}{12} + \frac{1}{12}

Solution steps

y^{'} (x) + 5 y x = \frac{x}{2} - x y

Solve separable ODE:

y = \frac{e ^{- 3 x^{2} + c_{1}}}{12} + \frac{1}{12}

y = \frac{e ^{- 3 x^{2} + c_{1}}}{12} + \frac{1}{12}

\frac{d y}{d x} = 12 x^{3} e^{y}

\frac{d x}{d t} = - \frac{x}{t}

\frac{d y}{d x} = \frac{x ^{2} + 1}{2 - y}

\frac{d y}{d x} = \frac{( x - 1 ) ^{2}}{3 y}

\frac{d x}{x ^{2} y} = \frac{d y}{2 y ^{2}}

What is the solution for y^'+5yx= x/2-xy ?
The solution for y^'+5yx= x/2-xy is y=(e^{-3x^2+c_{1}})/(12)+1/12