What is the solution for x^2y^'-2-3y=0 ?

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

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x^2y^'-2-3y=0

x^{2} y^{^{'}} - 2 - 3 y = 0

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

Solution steps

x^{2} y^{'} (x) - 2 - 3 y = 0

Solve separable ODE:

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

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

y^{^{'}} + x^{2} y = x^{2} y^{- 3}

(1 - 3 y^{2}) d y - x sin (x) d x = 0, y (\frac{π}{2}) = 2

\frac{d y}{d x} = 6 x 25 - y^{2}

3 x y y^{^{'}} + 4 y^{1} - 2 = 0

\frac{d y}{d t} + 1 + t^{2} y = 0, y (0) = 5

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