What is the solution for (dy)/(dt)=-ky+100k,y(0)=50 ?

The solution for (dy)/(dt)=-ky+100k,y(0)=50 is y=100-(50)/(e^{kt)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dy)/(dt)=-ky+100k,y(0)=50

\frac{d y}{d t} = - k y + 100 k, y (0) = 50

y = 100 - \frac{50}{e ^{k t}}

Solution steps

\frac{d y}{d t} = - k y (t) + 100 k

Solve linear ODE:

y = 100 - \frac{50}{e ^{k t}}

y = 100 - \frac{50}{e ^{k t}}

(2 ln (y) + x ln (y)) d x + x y^{- 1} d y = 0

t x^{^{'}} = - t^{2} x^{3}

x^{^{'}} = x (t^{2} - 1) (t^{2} - 9), x (0) = 1

y^{^{'}} = \frac{t ( 4 y - y ^{2} )}{3}

\frac{d y}{d x} y = ln (2 - x - x^{2})

What is the solution for (dy)/(dt)=-ky+100k,y(0)=50 ?
The solution for (dy)/(dt)=-ky+100k,y(0)=50 is y=100-(50)/(e^{kt)}