What is the solution for (dy)/(dx)=e^y(2x^3-4x) ?

The solution for (dy)/(dx)=e^y(2x^3-4x) is y=-ln(-(x^4)/2+2x^2-c_{1})

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(dy)/(dx)=e^y(2x^3-4x)

\frac{d y}{d x} = e^{y} (2 x^{3} - 4 x)

y = - ln (- \frac{x ^{4}}{2} + 2 x^{2} - c_{1})

Solution steps

\frac{d y}{d x} = e^{y} (2 x^{3} - 4 x)

Solve separable ODE:

y = - ln (- \frac{x ^{4}}{2} + 2 x^{2} - c_{1})

y = - ln (- \frac{x ^{4}}{2} + 2 x^{2} - c_{1})

\int (\frac{x ^{3} + 2 x ^{2} - 4}{x ^{2}}) d x

\int_{1}^{2} x - 1 d x

\frac{3 s}{( s ^{2} + 4 )}

y^{^{''''}} + 2 y^{^{''}} + y = 0

\frac{d}{d x} ((\frac{10}{4 + x})^{\frac{1}{2}})

What is the solution for (dy)/(dx)=e^y(2x^3-4x) ?
The solution for (dy)/(dx)=e^y(2x^3-4x) is y=-ln(-(x^4)/2+2x^2-c_{1})