What is the solution for y^{''}+20y^'=0,y(0)=-5,y^'(0)=20 ?

The solution for y^{''}+20y^'=0,y(0)=-5,y^'(0)=20 is y=-e^{-20t}-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+20y^'=0,y(0)=-5,y^'(0)=20

y^{^{''}} + 20 y^{^{'}} = 0, y (0) = - 5, y^{^{'}} (0) = 20

y = - e^{- 20 t} - 4

Solution steps

y^{''} (t) + 20 y^{'} (t) = 0

Solve linear ODE:

y = - e^{- 20 t} - 4

y = - e^{- 20 t} - 4

\frac{d ^{2}}{d x ^{2}} (x^{\frac{1}{11}})

\frac{\partial}{\partial x} (x e^{x^{2} + y^{2}})

\frac{\partial}{\partial x} (\frac{e ^{v}}{u ( v , x ) + v ^{2}})

(e^{x})^{^{'}}

f (x) = x^{5} + \frac{1}{x ^{5}}

What is the solution for y^{''}+20y^'=0,y(0)=-5,y^'(0)=20 ?
The solution for y^{''}+20y^'=0,y(0)=-5,y^'(0)=20 is y=-e^{-20t}-4