What is the solution for y^'=-0.2y,y(0)=ce^{0.2t} ?

The solution for y^'=-0.2y,y(0)=ce^{0.2t} is y=e^{-0.2t+c_{1}}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=-0.2y,y(0)=ce^{0.2t}

y^{^{'}} = - 0.2 y, y (0) = c e^{0.2 t}

y = e^{- 0.2 t + c_{1}}

Solution steps

y^{'} (t) = - 0.2 y

Solve separable ODE:

y = e^{- 0.2 t + c_{1}}

y = e^{- 0.2 t + c_{1}}

\frac{d y}{d x} + 12 x^{2} y = 7 x^{2}, y (0) = 2

\frac{d y}{d x} = - y x

y^{^{'}} + 9 x y = 8 x, y (0) = 6

\frac{d y}{d x} = \frac{3 x ^{2} + 4 x + 2}{2 y + 1}

x^{2} + y^{2} \frac{d y}{d x} = 0

What is the solution for y^'=-0.2y,y(0)=ce^{0.2t} ?
The solution for y^'=-0.2y,y(0)=ce^{0.2t} is y=e^{-0.2t+c_{1}}