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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

y^{^{''}} - 5 y^{^{'}} + 6 y = - sin (5 t)

\frac{d y}{d x} = \frac{1}{e ^{y} - x}

\frac{d y}{d x}, \frac{6 x}{y ^{7}} = 4

\frac{\partial}{\partial t} (e^{t} cos (x))

20 t e^{- 2 t} \cdot sin (4 t)

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