What is the solution for x^{''}-6x^'+25x=64e^{-t} ?

The solution for x^{''}-6x^'+25x=64e^{-t} is x=e^{3t}(c_{1}cos(4t)+c_{2}sin(4t))+2e^{-t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{''}-6x^'+25x=64e^{-t}

x^{^{''}} - 6 x^{^{'}} + 25 x = 64 e^{- t}

x = e^{3 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + 2 e^{- t}

Solution steps

x^{''} (t) - 6 x^{'} (t) + 25 x = 64 e^{- t}

Solve linear ODE:

x = e^{3 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + 2 e^{- t}

x = e^{3 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + 2 e^{- t}

\frac{d}{d x} (- 7 x^{3})

\frac{\partial}{\partial y} (ln (x^{2} + 1) + y^{2})

arctan (0)

x \to 1 lim (e^{x^{3} - x})

\frac{d}{d x} (\frac{3 x - 2}{x ^{2}})

What is the solution for x^{''}-6x^'+25x=64e^{-t} ?
The solution for x^{''}-6x^'+25x=64e^{-t} is x=e^{3t}(c_{1}cos(4t)+c_{2}sin(4t))+2e^{-t}