What is the solution for y^{''}-20y^'+116y=0 ?

The solution for y^{''}-20y^'+116y=0 is y=e^{10t}(c_{1}cos(4t)+c_{2}sin(4t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-20y^'+116y=0

y^{^{''}} - 20 y^{^{'}} + 116 y = 0

y = e^{10 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

Solution steps

y^{''} (t) - 20 y^{'} (t) + 116 y = 0

Solve linear ODE:

y = e^{10 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

y = e^{10 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

9 y^{^{''}} - 6 y^{^{'}} + y = 0, y (0) = 1, y^{^{'}} (0) = 2

2 y^{^{''}} + 12 y^{^{'}} + 19 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} + 32 y^{^{'}} + 160 y = 0

4 q^{^{''}} + 8 q^{^{'}} + \frac{1}{0.0125} q = 88 q

x^{^{''}} + 64 x = 0, x (0) = 0

What is the solution for y^{''}-20y^'+116y=0 ?
The solution for y^{''}-20y^'+116y=0 is y=e^{10t}(c_{1}cos(4t)+c_{2}sin(4t))