What is the solution for 8x^2y^'=y^'+8xe^{-y} ?

The solution for 8x^2y^'=y^'+8xe^{-y} is y=ln(1/2 ln(8x^2-1)+c_{1})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

8x^2y^'=y^'+8xe^{-y}

8 x^{2} y^{^{'}} = y^{^{'}} + 8 x e^{- y}

y = ln (\frac{1}{2} ln (8 x^{2} - 1) + c_{1})

Solution steps

8 x^{2} y^{'} (x) = y^{'} (x) + 8 x e^{- y}

Solve separable ODE:

y = ln (\frac{1}{2} ln (8 x^{2} - 1) + c_{1})

y = ln (\frac{1}{2} ln (8 x^{2} - 1) + c_{1})

f (x) = - 7 x - \frac{2}{x ^{7}}

\frac{\partial}{\partial x} (arctan (y))

x \to 2 lim (π)

\frac{d}{d x} (- sin (x) + cos (x))

f (x) = (6 x^{2} + 4) e^{x}

What is the solution for 8x^2y^'=y^'+8xe^{-y} ?
The solution for 8x^2y^'=y^'+8xe^{-y} is y=ln(1/2 ln(8x^2-1)+c_{1})