What is the solution for 8(dy)/(dx)+7e^{x+y}=0 ?

The solution for 8(dy)/(dx)+7e^{x+y}=0 is y=-ln((7e^x-c_{1})/8)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

8(dy)/(dx)+7e^{x+y}=0

8 \frac{d y}{d x} + 7 e^{x + y} = 0

y = - ln (\frac{7 e ^{x} - c _{1}}{8})

Solution steps

8 \frac{d y}{d x} + 7 e^{x + y} = 0

Solve separable ODE:

y = - ln (\frac{7 e ^{x} - c _{1}}{8})

y = - ln (\frac{7 e ^{x} - c _{1}}{8})

y^{^{'}} = t y

z \to 0 lim (\frac{1}{1 + z ^{2}})

\frac{d}{d x} (\frac{ln ( x ^{2} )}{x})

(t^{2} + 1) f^{^{'}} (t) + [f (t)]^{2} + 1 = 0

\int_{2}^{3} \frac{6}{3 - x} d x

What is the solution for 8(dy)/(dx)+7e^{x+y}=0 ?
The solution for 8(dy)/(dx)+7e^{x+y}=0 is y=-ln((7e^x-c_{1})/8)