What is the solution for linear x^{''}-(a-2)x^'-(a-1)x=0 ?

The solution for linear x^{''}-(a-2)x^'-(a-1)x=0 is x=c_{1}e^{(a-1)t}+c_{2}e^{-t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

linear x^{''}-(a-2)x^'-(a-1)x=0

linear

x^{^{''}} - (a - 2) x^{^{'}} - (a - 1) x = 0

x = c_{1} e^{(a - 1) t} + c_{2} e^{- t}

Solution steps

x^{''} (t) - (a - 2) x^{'} (t) - (a - 1) x = 0

Solve linear ODE:

x = c_{1} e^{(a - 1) t} + c_{2} e^{- t}

x = c_{1} e^{(a - 1) t} + c_{2} e^{- t}

3 y^{^{''}} - 6 y^{^{'}} + 9 y = 0

y^{^{''}} + 25 y = 0, y (0) = 4, y^{^{'}} (0) = - 2

x^{''} + x = 0, x (\frac{π}{2}) = 0, x^{^{'}} (\frac{π}{2}) = 1

y^{^{'''''}} + 3 y^{^{''''}} + 4 y^{^{'''}} + 12 y^{^{''}} = 0

y^{^{'''}} - 2 y^{^{''}} - 13 y^{^{'}} - 10 y = 0

What is the solution for linear x^{''}-(a-2)x^'-(a-1)x=0 ?
The solution for linear x^{''}-(a-2)x^'-(a-1)x=0 is x=c_{1}e^{(a-1)t}+c_{2}e^{-t}