What is the solution for 2xdx-ydy=0 ?

The solution for 2xdx-ydy=0 is y=sqrt(2x^2+c_{1)},y=-sqrt(2x^2+c_{1)}

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2xdx-ydy=0

2 x d x - y d y = 0

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

Solution steps

2 x d x - y d y = 0

Solve separable ODE:

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

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

\frac{d y}{d x} = \frac{20 x ^{4}}{y}, y > 0, x > 0

\frac{d y}{d x} = \frac{x ^{2} e ^{x^{3} - y^{2}}}{y}

\frac{d y}{d x} - x^{3} = x^{3} y

y^{^{'}} = \frac{5 x ^{2} - 1}{5 + 11 y}

y^{^{'}} = y^{2}, y (1) = 1

What is the solution for 2xdx-ydy=0 ?
The solution for 2xdx-ydy=0 is y=sqrt(2x^2+c_{1)},y=-sqrt(2x^2+c_{1)}