What is the solution for xy^'+y=(xy)^{3/2} ?

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

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xy^'+y=(xy)^{3/2}

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

x \to 0 lim (\frac{x}{x ^{2} - 9 x})

x^{2} y^{^{'}} + 2 x y = ln (x), y (1) = 3

\int \frac{( 2 e ^{x} )}{x} d x

\frac{d}{d x} (cos (2 x) - 2 sin (x))

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

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