What is the solution for dy=sqrt(e^{2y)}+1dx ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

dy=sqrt(e^{2y)}+1dx

d y = e^{2 y} + 1 d x

y = - W (- \frac{1}{e ^{c_{1} - x}}) + x - c_{1}

Solution steps

d y = e^{2 y} + 1 d x

Let

x

be the dependent variable. Divide by

d y

1 = e^{2 y} + 1 \cdot \frac{d x}{d y}

Substitute

\frac{d x}{d y}

with

x^{^{'}}

1 = e^{2 y} + 1 x^{^{^{'}}}

Isolate

x^{^{'}} : x^{^{'}} = 1 - e^{y}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

x = \int 1 - e^{y} d y

\int 1 - e^{y} d y = y - e^{y} + c_{1}

x = y - e^{y} + c_{1}

Isolate

y : y = - W (- \frac{1}{e ^{c_{1} - x}}) + x - c_{1}

y = - W (- \frac{1}{e ^{c_{1} - x}}) + x - c_{1}

\frac{d y}{d x} = e^{x} sin (2 y)

\frac{d y}{d x} = \frac{x}{y}, y (3) = 5

y^{\frac{3}{2}} \frac{d y}{d x} = 1

\frac{d y}{d x} y = csc^{2} (x)

\frac{d y}{d x} = y^{- 2} ln (x)

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