What is the solution for y^{''}+2y^'+y=0,y(0)=-4,y^'(0)=-1 ?

The solution for y^{''}+2y^'+y=0,y(0)=-4,y^'(0)=-1 is y=-4e^{-t}-5e^{-t}t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'+y=0,y(0)=-4,y^'(0)=-1

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = - 4, y^{^{'}} (0) = - 1

y = - 4 e^{- t} - 5 e^{- t} t

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = 0

Solve linear ODE:

y = - 4 e^{- t} - 5 e^{- t} t

y = - 4 e^{- t} - 5 e^{- t} t

e^{x} - 1

\int_{1}^{\infty} x^{- \frac{5}{4}} d x

\int \frac{1}{x + x} d x

\frac{\partial}{\partial x} (\frac{x ^{2} - y ^{2}}{z ^{2}})

\frac{d}{d x} (x^{3} - \frac{3}{2} x^{2})

What is the solution for y^{''}+2y^'+y=0,y(0)=-4,y^'(0)=-1 ?
The solution for y^{''}+2y^'+y=0,y(0)=-4,y^'(0)=-1 is y=-4e^{-t}-5e^{-t}t