What is the solution for x^{''}+4x^'+3x=0,x(0)=2 ?

The solution for x^{''}+4x^'+3x=0,x(0)=2 is x=(2-c_{2})e^{-t}+c_{2}e^{-3t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{''}+4x^'+3x=0,x(0)=2

x^{^{''}} + 4 x^{^{'}} + 3 x = 0, x (0) = 2

x = (2 - c_{2}) e^{- t} + c_{2} e^{- 3 t}

Solution steps

x^{''} (t) + 4 x^{'} (t) + 3 x = 0

Solve linear ODE:

x = (2 - c_{2}) e^{- t} + c_{2} e^{- 3 t}

x = (2 - c_{2}) e^{- t} + c_{2} e^{- 3 t}

\frac{3}{4} \cdot x^{^{''}} + 36 x = 0, x^{^{'}} (0) = 0, x (0) = - \frac{2}{3}

y^{^{''}} - 6 y^{^{'}} - 12 y = 0

y^{^{'}} + 3 y^{^{''}} = 0

y^{^{''}} - 8 y^{^{'}} - 105 y = 0

\frac{d ^{2} x}{d t ^{2}} + \frac{d x}{d t} + x = 0

What is the solution for x^{''}+4x^'+3x=0,x(0)=2 ?
The solution for x^{''}+4x^'+3x=0,x(0)=2 is x=(2-c_{2})e^{-t}+c_{2}e^{-3t}