What is the solution for (1xy^2+4y)dx+(1x^2y+4x)dy=0 ?

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

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(1xy^2+4y)dx+(1x^2y+4x)dy=0

(1 x y^{2} + 4 y) d x + (1 x^{2} y + 4 x) d y = 0

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

Solution steps

(1 x y^{2} + 4 y) d x + (1 x^{2} y + 4 x) d y = 0

Solve linear ODE:

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

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

y = x^{2} - 6 x + 9

\frac{d}{d x} (a x^{2} + b x)

\int_{1}^{2} (\frac{1}{8} + \frac{3}{8} x) d x

f (x) = \frac{cos ( x )}{x ^{4}}

x \to - 2 - lim (\frac{x - 3}{x ^{2} - 4})

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