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

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

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2xdx-ydy=0,y=f

2 x d x - y d y = 0, y = f

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} 4 x^{3} y^{4} = 8

\frac{d y}{d t} = - 0.17 (100 - y)

\frac{( d y )}{( d x )} = \frac{( 1 )}{( 4 ) y}

y^{^{'}} = - 2 y - 8, y (0) = 0

y^{^{'}} = - \frac{3 y}{1000 + 3 t}

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