What is the solution for y^{''}+7y^'+12y=0,y(0)=a,y^'(0)=b ?

The solution for y^{''}+7y^'+12y=0,y(0)=a,y^'(0)=b is y=(a-(-b-3a))e^{-3t}+(-b-3a)e^{-4t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+7y^'+12y=0,y(0)=a,y^'(0)=b

y^{^{''}} + 7 y^{^{'}} + 12 y = 0, y (0) = a, y^{^{'}} (0) = b

y = (a - (- b - 3 a)) e^{- 3 t} + (- b - 3 a) e^{- 4 t}

Solution steps

y^{''} (t) + 7 y^{'} (t) + 12 y = 0

Solve linear ODE:

y = (a - (- b - 3 a)) e^{- 3 t} + (- b - 3 a) e^{- 4 t}

y = (a - (- b - 3 a)) e^{- 3 t} + (- b - 3 a) e^{- 4 t}

\int_{- 1}^{2} x^{4} d x

n \to \infty lim (D n)

\int cos^{6} (x) tan^{3} (x) d x

\frac{\partial}{\partial x} (sec (x) tan (y))

\int cos^{2} (x) + sin^{2} (x) d x

What is the solution for y^{''}+7y^'+12y=0,y(0)=a,y^'(0)=b ?
The solution for y^{''}+7y^'+12y=0,y(0)=a,y^'(0)=b is y=(a-(-b-3a))e^{-3t}+(-b-3a)e^{-4t}