What is the solution for y^{''}-4y^'+y5=0 ?

The solution for y^{''}-4y^'+y5=0 is y=e^{2t}(c_{1}cos(t)+c_{2}sin(t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-4y^'+y5=0

y^{^{''}} - 4 y^{^{'}} + y 5 = 0

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

Solution steps

y^{''} (t) - 4 y^{'} (t) + y 5 = 0

Solve linear ODE:

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

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

y^{^{''}} - 9 y^{^{'}} + 7 y = 0

3 y^{^{'''}} + 5 y^{^{''}} = 0

5 y^{^{''}} + 5 y^{^{'}} + 10 y = 0

y^{^{''}} + 2 y^{^{'}} - 8 y = 0, y (0) = - 2, y^{^{'}} (0) = 1

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

What is the solution for y^{''}-4y^'+y5=0 ?
The solution for y^{''}-4y^'+y5=0 is y=e^{2t}(c_{1}cos(t)+c_{2}sin(t))