What is the solution for x^{''}+100x^'+2500x=30 ?

The solution for x^{''}+100x^'+2500x=30 is x=c_{1}e^{-50t}+c_{2}te^{-50t}+3/250

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{''}+100x^'+2500x=30

x^{^{''}} + 100 x^{^{'}} + 2500 x = 30

x = c_{1} e^{- 50 t} + c_{2} t e^{- 50 t} + \frac{3}{250}

Solution steps

x^{''} (t) + 100 x^{'} (t) + 2500 x = 30

Solve linear ODE:

x = c_{1} e^{- 50 t} + c_{2} t e^{- 50 t} + \frac{3}{250}

x = c_{1} e^{- 50 t} + c_{2} t e^{- 50 t} + \frac{3}{250}

16 y^{^{''}} + 24 y^{^{'}} + 9 = 0

(D^{3} + D^{2} - 4 D - 4) y = 3 e^{- x} - 4 x - 6

y^{^{''}} + 2 y^{^{'}} + 5 y = 30 x^{3} + 2

y^{^{''}} + 4 y^{^{'}} + 5 y = 2 e^{x}

y^{^{''}} + 6 y^{^{'}} + 9 y = 12 e^{t}, y (0) = - 1, y^{^{'}} (0) = 7

What is the solution for x^{''}+100x^'+2500x=30 ?
The solution for x^{''}+100x^'+2500x=30 is x=c_{1}e^{-50t}+c_{2}te^{-50t}+3/250