What is the solution for y^'+1/(200+t)y=0,y(0)=60 ?

The solution for y^'+1/(200+t)y=0,y(0)=60 is y=(12000)/(t+200)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+1/(200+t)y=0,y(0)=60

y^{^{'}} + \frac{1}{200 + t} y = 0, y (0) = 60

y = \frac{12000}{t + 200}

Solution steps

y^{'} (t) + \frac{1}{200 + t} y = 0

Solve separable ODE:

y = \frac{12000}{t + 200}

y = \frac{12000}{t + 200}

\int \frac{4 x ^{2} sin ( 2 x )}{9} d x

n = 2 \sum \infty \frac{n - 2}{n 5 ^{n}}

n = 3 \sum \infty \frac{7 ^{n}}{1 4 ^{n}}

x \to \infty lim (2 + \frac{5}{x})

\int \frac{e ^{- x^{2}}}{x} d x

What is the solution for y^'+1/(200+t)y=0,y(0)=60 ?
The solution for y^'+1/(200+t)y=0,y(0)=60 is y=(12000)/(t+200)