What is the solution for 25x^{''}+50x^'+169x=0,x(0)=2,x^'(0)=6 ?

The solution for 25x^{''}+50x^'+169x=0,x(0)=2,x^'(0)=6 is x=e^{-t}(2cos((12t)/5)+10/3 sin((12t)/5))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

25x^{''}+50x^'+169x=0,x(0)=2,x^'(0)=6

25 x^{^{''}} + 50 x^{^{'}} + 169 x = 0, x (0) = 2, x^{^{'}} (0) = 6

x = e^{- t} (2 cos (\frac{12 t}{5}) + \frac{10}{3} sin (\frac{12 t}{5}))

Solution steps

25 x^{''} (t) + 50 x^{'} (t) + 169 x = 0

Solve linear ODE:

x = e^{- t} (2 cos (\frac{12 t}{5}) + \frac{10}{3} sin (\frac{12 t}{5}))

x = e^{- t} (2 cos (\frac{12 t}{5}) + \frac{10}{3} sin (\frac{12 t}{5}))

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = k, y^{^{'}} (0) = 2

y^{^{''}} + 2 y = 0, y (π) = 0, y (0) = 1

3 y^{^{''}} - 15 y^{^{'}} + 4 y = 0

1 y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 12 y^{^{'}} + 27 y = 0, y (0) = 1, y^{^{'}} (0) = - 12

What is the solution for 25x^{''}+50x^'+169x=0,x(0)=2,x^'(0)=6 ?
The solution for 25x^{''}+50x^'+169x=0,x(0)=2,x^'(0)=6 is x=e^{-t}(2cos((12t)/5)+10/3 sin((12t)/5))