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}

\int_{- 9}^{0} 2 + 81 - x^{2} d x

\int \frac{- x - 1}{x ^{2} + 1} d x

\frac{\partial}{\partial y} (2 x + y^{2})

f (x) = x^{3}, 0, 4

y^{^{'}} - 3 y = - 6 y^{3}

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}