What is the solution for (2x^3-1)dx-ydy=0 ?

The solution for (2x^3-1)dx-ydy=0 is y=sqrt(x^4-2x-c_{1)},y=-sqrt(x^4-2x-c_{1)}

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(2x^3-1)dx-ydy=0

(2 x^{3} - 1) d x - y d y = 0

y = x^{4} - 2 x - c_{1}, y = - x^{4} - 2 x - c_{1}

Solution steps

(2 x^{3} - 1) d x - y d y = 0

Solve separable ODE:

y = x^{4} - 2 x - c_{1}, y = - x^{4} - 2 x - c_{1}

y = x^{4} - 2 x - c_{1}, y = - x^{4} - 2 x - c_{1}

(x - 1) d x - (x e^{y} + e^{y}) d y = 0

y^{^{'}} = k \frac{x}{y}

\frac{d y}{d t} - y = 0, y (0) = 1

\frac{d y}{d x} = \frac{y - 1}{x ^{2} 4 + x ^{2}}

\frac{d P}{d t} = 7 Pt

What is the solution for (2x^3-1)dx-ydy=0 ?
The solution for (2x^3-1)dx-ydy=0 is y=sqrt(x^4-2x-c_{1)},y=-sqrt(x^4-2x-c_{1)}