What is the solution for y^'+y(3/x+x/2)=0 ?

The solution for y^'+y(3/x+x/2)=0 is y=(e^{(-x^2+c_{1})/4})/(x^3)

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y^'+y(3/x+x/2)=0

y^{^{'}} + y (\frac{3}{x} + \frac{x}{2}) = 0

y = \frac{e ^{\frac{- x ^{2} + c _{1}}{4}}}{x ^{3}}

Solution steps

y^{'} (x) + y (\frac{3}{x} + \frac{x}{2}) = 0

Solve separable ODE:

y = \frac{e ^{\frac{- x ^{2} + c _{1}}{4}}}{x ^{3}}

y = \frac{e ^{\frac{- x ^{2} + c _{1}}{4}}}{x ^{3}}

f (x) = \frac{1}{7 x - 4}

x^{3} + x + 12 x^{\frac{1}{2}}

\frac{d}{d x} ((5 x + 1)^{3} (6 x^{2} + 3 x + 2)^{5})

\frac{d}{d x} (\frac{sec ^{2} ( x ^{3} )}{2})^{4}

\frac{d}{d x} (4 x (3 x - 2)^{5})

What is the solution for y^'+y(3/x+x/2)=0 ?
The solution for y^'+y(3/x+x/2)=0 is y=(e^{(-x^2+c_{1})/4})/(x^3)