What is the solution for 8x^2y^'=y^'+5xe^{-y} ?

The solution for 8x^2y^'=y^'+5xe^{-y} is y=ln(5/16 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^'+5xe^{-y}

8 x^{2} y^{^{'}} = y^{^{'}} + 5 x e^{- y}

y = ln (\frac{5}{16} ln (8 x^{2} - 1) + c_{1})

Solution steps

8 x^{2} y^{'} (x) = y^{'} (x) + 5 x e^{- y}

Solve separable ODE:

y = ln (\frac{5}{16} ln (8 x^{2} - 1) + c_{1})

y = ln (\frac{5}{16} ln (8 x^{2} - 1) + c_{1})

\frac{d y}{d x} + y^{3} x + 2 y = 0

\int \frac{2}{x} + \frac{x}{2} d x

x \to \frac{π}{2} lim (\frac{csc ( 4 x )}{csc ( 2 x )})

f (x) = \frac{2 x - 1}{x + 5}, at x = 1

x \to - \infty lim (\frac{4 x ^{6}}{x ^{3} - 8})

What is the solution for 8x^2y^'=y^'+5xe^{-y} ?
The solution for 8x^2y^'=y^'+5xe^{-y} is y=ln(5/16 ln(8x^2-1)+c_{1})