What is the solution for x^{''}-26x^'+25x=0,x(0)=-1,x^'(0)=2 ?

The solution for x^{''}-26x^'+25x=0,x(0)=-1,x^'(0)=2 is x=(-1+9/8)e^{25t}-9/8 e^t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{''}-26x^'+25x=0,x(0)=-1,x^'(0)=2

x^{^{''}} - 26 x^{^{'}} + 25 x = 0, x (0) = - 1, x^{^{'}} (0) = 2

x = (- 1 + \frac{9}{8}) e^{25 t} - \frac{9}{8} e^{t}

Solution steps

x^{''} (t) - 26 x^{'} (t) + 25 x = 0

Solve linear ODE:

x = (- 1 + \frac{9}{8}) e^{25 t} - \frac{9}{8} e^{t}

x = (- 1 + \frac{9}{8}) e^{25 t} - \frac{9}{8} e^{t}

D^{2} (D + 4)^{3} y = 0

y^{^{'''}} - 8 y^{^{''}} + 15 y^{^{'}} = 0

y^{^{'''}} + 15 y^{^{''}} + 58 y^{^{'}} + 44 y = 0

y^{^{''}} - 6 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} - 2 y^{^{'}} = 0, y (0) = 3 - 3 e^{2}, y (1) = 0

What is the solution for x^{''}-26x^'+25x=0,x(0)=-1,x^'(0)=2 ?
The solution for x^{''}-26x^'+25x=0,x(0)=-1,x^'(0)=2 is x=(-1+9/8)e^{25t}-9/8 e^t