What is the solution for (8+x)y^'=(x+9)(y-8) ?

The solution for (8+x)y^'=(x+9)(y-8) is y=8e^{x+c_{1}}+e^{x+c_{1}}x+8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(8+x)y^'=(x+9)(y-8)

(8 + x) y^{^{'}} = (x + 9) (y - 8)

y = 8 e^{x + c_{1}} + e^{x + c_{1}} x + 8

Solution steps

(8 + x) y^{'} (x) = (x + 9) (y - 8)

Solve separable ODE:

y = 8 e^{x + c_{1}} + e^{x + c_{1}} x + 8

y = 8 e^{x + c_{1}} + e^{x + c_{1}} x + 8

y^{2} d x - x y d y = x^{2} y d y, y (1) = 1

\frac{d y}{d x} = x (1 + y^{2}), y (0) = 0

\frac{d y}{d x} 5 x + 7 y + 6 = 0

y^{^{'}} = 1 + x + y^{2} + x y^{2}, y (0) = 1

\frac{d x}{d t} = k (a - x)

What is the solution for (8+x)y^'=(x+9)(y-8) ?
The solution for (8+x)y^'=(x+9)(y-8) is y=8e^{x+c_{1}}+e^{x+c_{1}}x+8