What is the solution for y^{''}+8y^'+17y=0,y(0)=1,y^'(0)=-5 ?

The solution for y^{''}+8y^'+17y=0,y(0)=1,y^'(0)=-5 is y=e^{-4t}(cos(t)-sin(t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+8y^'+17y=0,y(0)=1,y^'(0)=-5

y^{^{''}} + 8 y^{^{'}} + 17 y = 0, y (0) = 1, y^{^{'}} (0) = - 5

y = e^{- 4 t} (cos (t) - sin (t))

Solution steps

y^{''} (t) + 8 y^{'} (t) + 17 y = 0

Solve linear ODE:

y = e^{- 4 t} (cos (t) - sin (t))

y = e^{- 4 t} (cos (t) - sin (t))

9 y - y^{^{''}} = 0

y^{^{''}} - 10 y^{^{'}} + 34 y = 0, y (\frac{π}{6}) = 1, y (0) = 4

10 y^{^{''}} + 30 y^{^{'}} + 20 y = 0, y (0) = - 7, y^{^{'}} (0) = - 2

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

y^{^{''}} - 4 \cdot y^{^{'}} + 4 \cdot y = 0

What is the solution for y^{''}+8y^'+17y=0,y(0)=1,y^'(0)=-5 ?
The solution for y^{''}+8y^'+17y=0,y(0)=1,y^'(0)=-5 is y=e^{-4t}(cos(t)-sin(t))