What is the solution for y^{''}+12y^'+36y=0,y(0)=1,y^'(0)=0 ?

The solution for y^{''}+12y^'+36y=0,y(0)=1,y^'(0)=0 is y=e^{-6t}+6e^{-6t}t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+12y^'+36y=0,y(0)=1,y^'(0)=0

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y = e^{- 6 t} + 6 e^{- 6 t} t

Solution steps

y^{''} (t) + 12 y^{'} (t) + 36 y = 0

Solve linear ODE:

y = e^{- 6 t} + 6 e^{- 6 t} t

y = e^{- 6 t} + 6 e^{- 6 t} t

\frac{d}{d x} (tan (e^{π}))

\frac{d}{d x} (\frac{1}{2 x + 6})

\int \frac{1}{4 x ^{2}} d x

\int \frac{- 8 x ^{2}}{4 x - x ^{2}} d x

\frac{d}{d x} (3 cos (x) sin (x))

What is the solution for y^{''}+12y^'+36y=0,y(0)=1,y^'(0)=0 ?
The solution for y^{''}+12y^'+36y=0,y(0)=1,y^'(0)=0 is y=e^{-6t}+6e^{-6t}t