What is the solution for (dy)/(dx)=(((1+y)^2))/((1+x)^3) ?

The solution for (dy)/(dx)=(((1+y)^2))/((1+x)^3) is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dy)/(dx)=(((1+y)^2))/((1+x)^3)

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

y = - \frac{2 x ^{2}}{- 1 + 2 c _{1} x ^{2} + 4 c _{1} x + 2 c _{1}} - \frac{2 c _{1} x ^{2}}{- 1 + 2 c _{1} x ^{2} + 4 c _{1} x + 2 c _{1}} - \frac{4 x}{- 1 + 2 c _{1} x ^{2} + 4 c _{1} x + 2 c _{1}} - \frac{4 c _{1} x}{- 1 + 2 c _{1} x ^{2} + 4 c _{1} x + 2 c _{1}} - \frac{2 c _{1}}{- 1 + 2 c _{1} x ^{2} + 4 c _{1} x + 2 c _{1}} - \frac{1}{- 1 + 2 c _{1} x ^{2} + 4 c _{1} x + 2 c _{1}}

Solution steps

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

Solve separable ODE:

y = - \frac{2 x ^{2}}{- 1 + 2 c _{1} x ^{2} + 4 c _{1} x + 2 c _{1}} - \frac{2 c _{1} x ^{2}}{- 1 + 2 c _{1} x ^{2} + 4 c _{1} x + 2 c _{1}} - \frac{4 x}{- 1 + 2 c _{1} x ^{2} + 4 c _{1} x + 2 c _{1}} - \frac{4 c _{1} x}{- 1 + 2 c _{1} x ^{2} + 4 c _{1} x + 2 c _{1}} - \frac{2 c _{1}}{- 1 + 2 c _{1} x ^{2} + 4 c _{1} x + 2 c _{1}} - \frac{1}{- 1 + 2 c _{1} x ^{2} + 4 c _{1} x + 2 c _{1}}

y = - \frac{2 x ^{2}}{- 1 + 2 c _{1} x ^{2} + 4 c _{1} x + 2 c _{1}} - \frac{2 c _{1} x ^{2}}{- 1 + 2 c _{1} x ^{2} + 4 c _{1} x + 2 c _{1}} - \frac{4 x}{- 1 + 2 c _{1} x ^{2} + 4 c _{1} x + 2 c _{1}} - \frac{4 c _{1} x}{- 1 + 2 c _{1} x ^{2} + 4 c _{1} x + 2 c _{1}} - \frac{2 c _{1}}{- 1 + 2 c _{1} x ^{2} + 4 c _{1} x + 2 c _{1}} - \frac{1}{- 1 + 2 c _{1} x ^{2} + 4 c _{1} x + 2 c _{1}}

\int e^{- 2 x} cos (e^{- x}) d x

\frac{d}{d y} (\frac{x + y}{x - y})

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

\int \frac{1}{2} sin (2 x) d x

F (t) = 2 + 4 e^{- 4 t} - 3 e^{3 t}

What is the solution for (dy)/(dx)=(((1+y)^2))/((1+x)^3) ?
The solution for (dy)/(dx)=(((1+y)^2))/((1+x)^3) is