What is the solution for 3xyy^'+4y^1-2=0 ?

The solution for 3xyy^'+4y^1-2=0 is y=(\W(1/(c_{1)x^{8/3}}))/2+1/2

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3xyy^'+4y^1-2=0

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

y = \frac{W ( \frac{1}{c _{1} x ^{\frac{8}{3}}} )}{2} + \frac{1}{2}

Solution steps

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

Solve separable ODE:

y = \frac{W ( \frac{1}{c _{1} x ^{\frac{8}{3}}} )}{2} + \frac{1}{2}

y = \frac{W ( \frac{1}{c _{1} x ^{\frac{8}{3}}} )}{2} + \frac{1}{2}

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

(x^{2} - y x^{2}) d y + (y^{2} + x y^{2}) d x = 0

d x + sin (t) x d t - sin (t) d t = 0, x (\frac{π}{2}) = - 1

y^{^{'}} + tan (3 x) y = 0

y^{^{'}} = e^{4 y + 10 x}

What is the solution for 3xyy^'+4y^1-2=0 ?
The solution for 3xyy^'+4y^1-2=0 is y=(\W(1/(c_{1)x^{8/3}}))/2+1/2