What is the solution for (e^y-1)(dy)/(dx)=2+cos(x) ?

The solution for (e^y-1)(dy)/(dx)=2+cos(x) is y=-\W(-1/(e^{c_{1)+2x+sin(x)}})-2x-sin(x)-c_{1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(e^y-1)(dy)/(dx)=2+cos(x)

(e^{y} - 1) \frac{d y}{d x} = 2 + cos (x)

y = - W (- \frac{1}{e ^{c_{1} + 2 x + s i n (x)}}) - 2 x - sin (x) - c_{1}

Solution steps

(e^{y} - 1) \frac{d y}{d x} = 2 + cos (x)

Solve separable ODE:

y = - W (- \frac{1}{e ^{c_{1} + 2 x + s i n (x)}}) - 2 x - sin (x) - c_{1}

y = - W (- \frac{1}{e ^{c_{1} + 2 x + s i n (x)}}) - 2 x - sin (x) - c_{1}

y^{^{'}} + 5 y^{3} = 0

\frac{d y}{d x} = \frac{x ^{2} + 1}{y ^{2} + 1}, y (2) = 2

y^{'} (t) = y (y + 1)^{2}, y (0) = 1

\frac{d x}{d t} = x^{2} e^{- 2 t}

\frac{d y}{d x} = \frac{35}{y ^{\frac{1}{6}} + 49 x ^{2} y ^{\frac{1}{6}}}

What is the solution for (e^y-1)(dy)/(dx)=2+cos(x) ?
The solution for (e^y-1)(dy)/(dx)=2+cos(x) is y=-\W(-1/(e^{c_{1)+2x+sin(x)}})-2x-sin(x)-c_{1}