What is the solution for (x+(2y^2)/x)dx=ydy ?

The solution for (x+(2y^2)/x)dx=ydy is y=xsqrt(-1+c_{1)x^2},y=-xsqrt(-1+c_{1)x^2}

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(x+(2y^2)/x)dx=ydy

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

y = x - 1 + c_{1} x^{2}, y = - x - 1 + c_{1} x^{2}

Solution steps

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

Solve Bernoulli ODE:

y = x - 1 + c_{1} x^{2}, y = - x - 1 + c_{1} x^{2}

y = x - 1 + c_{1} x^{2}, y = - x - 1 + c_{1} x^{2}

(\frac{e ^{x}}{x})^{^{'}}

f (t) = (3 t + 14)^{\frac{2}{3}}

\frac{d y}{d x} + \frac{y}{x ^{2} + x} = x + 1

\frac{d}{d x} (2 I n x)

\frac{d}{d x} (3 e^{- 4 x})

What is the solution for (x+(2y^2)/x)dx=ydy ?
The solution for (x+(2y^2)/x)dx=ydy is y=xsqrt(-1+c_{1)x^2},y=-xsqrt(-1+c_{1)x^2}