What is the solution for y^'+y= 1/(1+e^t) ?

The solution for y^'+y= 1/(1+e^t) is y=(ln(1+e^t))/(e^t)+(c_{1})/(e^t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+y= 1/(1+e^t)

y^{^{'}} + y = \frac{1}{1 + e ^{t}}

y = \frac{ln ( 1 + e ^{t} )}{e ^{t}} + \frac{c _{1}}{e ^{t}}

Solution steps

y^{'} (t) + y = \frac{1}{1 + e ^{t}}

Solve linear ODE:

y = \frac{ln ( 1 + e ^{t} )}{e ^{t}} + \frac{c _{1}}{e ^{t}}

y = \frac{ln ( 1 + e ^{t} )}{e ^{t}} + \frac{c _{1}}{e ^{t}}

y^{^{'}} + 7 x^{6} y = x^{6}

x \to 0 lim (x^{2})

x \to 1 lim (6 x)

y = - 4 - 7 x^{2}, (- 4, - 16)

2 x^{5} ln (x)

What is the solution for y^'+y= 1/(1+e^t) ?
The solution for y^'+y= 1/(1+e^t) is y=(ln(1+e^t))/(e^t)+(c_{1})/(e^t)