What is the solution for y^{''}+20y^'+36y=0,y(0)=3,y^'(0)=0 ?

The solution for y^{''}+20y^'+36y=0,y(0)=3,y^'(0)=0 is y=(3+3/8)e^{-2t}-3/8 e^{-18t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+20y^'+36y=0,y(0)=3,y^'(0)=0

y^{^{''}} + 20 y^{^{'}} + 36 y = 0, y (0) = 3, y^{^{'}} (0) = 0

y = (3 + \frac{3}{8}) e^{- 2 t} - \frac{3}{8} e^{- 18 t}

Solution steps

y^{''} (t) + 20 y^{'} (t) + 36 y = 0

Solve linear ODE:

y = (3 + \frac{3}{8}) e^{- 2 t} - \frac{3}{8} e^{- 18 t}

y = (3 + \frac{3}{8}) e^{- 2 t} - \frac{3}{8} e^{- 18 t}

3 \frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + 3 y = 0

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

y^{^{'''}} - 7 y^{^{''}} + y^{^{'}} + 12 y = 0

y^{^{''}} - \frac{1}{4} y = 0, y (0) = 4, y^{^{'}} (0) = 0

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

What is the solution for y^{''}+20y^'+36y=0,y(0)=3,y^'(0)=0 ?
The solution for y^{''}+20y^'+36y=0,y(0)=3,y^'(0)=0 is y=(3+3/8)e^{-2t}-3/8 e^{-18t}