What is the solution for y^{''}-8y^'-7y=0 ?

The solution for y^{''}-8y^'-7y=0 is y=c_{1}e^{(4+sqrt(23))t}+c_{2}e^{(4-sqrt(23))t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-8y^'-7y=0

y^{^{''}} - 8 y^{^{'}} - 7 y = 0

y = c_{1} e^{(4 + 23) t} + c_{2} e^{(4 - 23) t}

Solution steps

y^{''} (t) - 8 y^{'} (t) - 7 y = 0

Solve linear ODE:

y = c_{1} e^{(4 + 23) t} + c_{2} e^{(4 - 23) t}

y = c_{1} e^{(4 + 23) t} + c_{2} e^{(4 - 23) t}

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

x^{^{''}} + 6 x^{^{'}} + 9 x = 0, x (0) = a, x^{^{'}} (0) = b

y^{^{''}} + 4 y^{^{'}} + 29 y = 0, y (0) = 3, y^{^{'}} (0) = - 2

9 (\frac{d ^{2} y}{d x ^{2}}) + 12 (\frac{d y}{d x}) + 4 y = 0

2 y^{^{''}} + 4 y^{^{'}} + k y = 0

What is the solution for y^{''}-8y^'-7y=0 ?
The solution for y^{''}-8y^'-7y=0 is y=c_{1}e^{(4+sqrt(23))t}+c_{2}e^{(4-sqrt(23))t}