What is the solution for y^{''}+8y^'+16y=0,y(0)=3,y^'(0)=8 ?

The solution for y^{''}+8y^'+16y=0,y(0)=3,y^'(0)=8 is y=3e^{-4t}+20te^{-4t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+8y^'+16y=0,y(0)=3,y^'(0)=8

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (0) = 3, y^{^{'}} (0) = 8

y = 3 e^{- 4 t} + 20 t e^{- 4 t}

Solution steps

y^{''} (t) + 8 y^{'} (t) + 16 y = 0

Solve linear ODE:

y = 3 e^{- 4 t} + 20 t e^{- 4 t}

y = 3 e^{- 4 t} + 20 t e^{- 4 t}

8 3 x^{2} + 9

\frac{\partial}{\partial x} (42 e^{7 x} y)

f (x) = \frac{x}{x ^{2} + 64}

\int_{0}^{1} 2 π (2 - y) (4 - 4 y^{2}) d y

y = - 2 x + 3, (0, 1)

What is the solution for y^{''}+8y^'+16y=0,y(0)=3,y^'(0)=8 ?
The solution for y^{''}+8y^'+16y=0,y(0)=3,y^'(0)=8 is y=3e^{-4t}+20te^{-4t}