en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'= 1/2 (k-y)

Solution

y^{^{'}} = \frac{1}{2} (k - y)

Solution

y = - e^{\frac{- t - c _{1}}{2}} + k

Solution steps

y^{'} (t) = \frac{1}{2} (k - y)

Solve separable ODE:

y = - e^{\frac{- t - c _{1}}{2}} + k

y = - e^{\frac{- t - c _{1}}{2}} + k

Graph

Popular Examples

(dy)/(dx)=6e^{2x-y},y(0)=0

\frac{d y}{d x} = 6 e^{2 x - y}, y (0) = 0

5xydx-ln^8(y)dy=0

5 x y d x - ln^{8} (y) d y = 0

y^{^{'}} = \frac{1}{3} y

(dy)/(dx)y=u^2-2u,u= 1/(x^2)

\frac{d y}{d x} y = u^{2} - 2 u, u = \frac{1}{x ^{2}}

3= pi/(12) d/(dt)((h(t))^3)

3 = \frac{π}{12} \frac{d}{d t} ((h (t))^{3})

Frequently Asked Questions (FAQ)

What is the solution for y^'= 1/2 (k-y) ?
The solution for y^'= 1/2 (k-y) is y=-e^{(-t-c_{1})/2}+k