What is the solution for (15xy^2-4y)+(15x^2y-4x)y^'=0 ?

The solution for (15xy^2-4y)+(15x^2y-4x)y^'=0 is y=(c_{1})/x

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(15xy^2-4y)+(15x^2y-4x)y^'=0

(15 x y^{2} - 4 y) + (15 x^{2} y - 4 x) y^{^{'}} = 0

y = \frac{c _{1}}{x}

Solution steps

(15 x y^{2} - 4 y) + (15 x^{2} y - 4 x) y^{'} (x) = 0

Solve linear ODE:

y = \frac{c _{1}}{x}

y = \frac{c _{1}}{x}

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\frac{d}{d t} (- \frac{1}{2 ( t + 1 ) \frac{3}{2}})

x \to - 2 + lim (\frac{3 x - 6}{x + 2})

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(x^{5} + y^{2}) d x - x y d y = 0

What is the solution for (15xy^2-4y)+(15x^2y-4x)y^'=0 ?
The solution for (15xy^2-4y)+(15x^2y-4x)y^'=0 is y=(c_{1})/x