What is the solution for y^{''}+8y^'+41y=0,y(0)=4,y^'(0)=-12 ?

The solution for y^{''}+8y^'+41y=0,y(0)=4,y^'(0)=-12 is y=e^{-4t}(4cos(5t)+4/5 sin(5t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+8y^'+41y=0,y(0)=4,y^'(0)=-12

y^{^{''}} + 8 y^{^{'}} + 41 y = 0, y (0) = 4, y^{^{'}} (0) = - 12

y = e^{- 4 t} (4 cos (5 t) + \frac{4}{5} sin (5 t))

Solution steps

y^{''} (t) + 8 y^{'} (t) + 41 y = 0

Solve linear ODE:

y = e^{- 4 t} (4 cos (5 t) + \frac{4}{5} sin (5 t))

y = e^{- 4 t} (4 cos (5 t) + \frac{4}{5} sin (5 t))

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

2 y^{^{''}} - 16 y^{^{'}} = 0

y^{^{''}} + 10 y^{^{'}} + 16 y = 0, y (0) = 1, y^{^{'}} (0) = - 12

y^{^{''}} + 9 π^{2} y = 0

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (- 1) = 2, y^{^{'}} (- 1) = 5

What is the solution for y^{''}+8y^'+41y=0,y(0)=4,y^'(0)=-12 ?
The solution for y^{''}+8y^'+41y=0,y(0)=4,y^'(0)=-12 is y=e^{-4t}(4cos(5t)+4/5 sin(5t))