What is the solution for y^'=(xy+3x-y-3)/(xy-2x+4y-8) ?

The solution for y^'=(xy+3x-y-3)/(xy-2x+4y-8) is y=-5\W(-(x+4}{5e^{\frac{x+c_{1}+3)/5}})-3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=(xy+3x-y-3)/(xy-2x+4y-8)

y^{^{'}} = \frac{x y + 3 x - y - 3}{x y - 2 x + 4 y - 8}

y = - 5 W (- \frac{x + 4}{5 e ^{\frac{x + c _{1} + 3}{5}}}) - 3

Solution steps

y^{'} (x) = \frac{x y + 3 x - y - 3}{x y - 2 x + 4 y - 8}

Solve separable ODE:

y = - 5 W (- \frac{x + 4}{5 e ^{\frac{x + c _{1} + 3}{5}}}) - 3

y = - 5 W (- \frac{x + 4}{5 e ^{\frac{x + c _{1} + 3}{5}}}) - 3

y = \frac{x ^{2} + 5}{x - 1}

\int (x^{2} + 6 x) cos (x) d x

\int \frac{8 + x + x ^{2}}{x} d x

\int_{0}^{\frac{1}{2}} 5 arccos (x) d x

\frac{d}{d x} (e^{a x^{6}})

What is the solution for y^'=(xy+3x-y-3)/(xy-2x+4y-8) ?
The solution for y^'=(xy+3x-y-3)/(xy-2x+4y-8) is y=-5\W(-(x+4}{5e^{\frac{x+c_{1}+3)/5}})-3