What is the solution for 4y^{''}+20y^'+169y=e^{2t} ?

The solution for 4y^{''}+20y^'+169y=e^{2t} is y=e^{-(5t)/2}(c_{1}cos(6t)+c_{2}sin(6t))+1/225 e^{2t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4y^{''}+20y^'+169y=e^{2t}

4 y^{^{''}} + 20 y^{^{'}} + 169 y = e^{2 t}

y = e^{- \frac{5 t}{2}} (c_{1} cos (6 t) + c_{2} sin (6 t)) + \frac{1}{225} e^{2 t}

Solution steps

4 y^{''} (t) + 20 y^{'} (t) + 169 y = e^{2 t}

Solve linear ODE:

y = e^{- \frac{5 t}{2}} (c_{1} cos (6 t) + c_{2} sin (6 t)) + \frac{1}{225} e^{2 t}

y = e^{- \frac{5 t}{2}} (c_{1} cos (6 t) + c_{2} sin (6 t)) + \frac{1}{225} e^{2 t}

(D^{3} + 8) y = x^{4} + 2 x + 1

y^{^{''}} - 6 y^{^{'}} + 13 y = 13

y^{^{''}} + 4 y = 16 t

y^{^{''}} - 8 y^{^{'}} + 16 y = e^{4 t} ln (t)

10 y^{^{''}} + 1000 y = 100 cos (8 t), y (0) = 0, y^{^{'}} (0) = 0

What is the solution for 4y^{''}+20y^'+169y=e^{2t} ?
The solution for 4y^{''}+20y^'+169y=e^{2t} is y=e^{-(5t)/2}(c_{1}cos(6t)+c_{2}sin(6t))+1/225 e^{2t}