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

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

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2xy(dy)/(dx)=x^2+2y^2

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

y = x ln (x) + c_{1}, y = - x ln (x) + c_{1}

Solution steps

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

Solve Bernoulli ODE:

y = x ln (x) + c_{1}, y = - x ln (x) + c_{1}

y = x ln (x) + c_{1}, y = - x ln (x) + c_{1}

\int sin (\frac{1}{x}) d x

f (x) = cot^{3} (x)

y = 2 x, y = 3 x - 4, y = 0

x \to 2 lim (\frac{2 x}{3 x + 1})

\frac{\partial}{\partial y} (y^{3} cos (2 x))

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