What is the solution for y^{''}-10y^'+50y=0 ?

The solution for y^{''}-10y^'+50y=0 is y=e^{5t}(c_{1}cos(5t)+c_{2}sin(5t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-10y^'+50y=0

y^{^{''}} - 10 y^{^{'}} + 50 y = 0

y = e^{5 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

Solution steps

y^{''} (t) - 10 y^{'} (t) + 50 y = 0

Solve linear ODE:

y = e^{5 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

y = e^{5 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

5 \cdot x^{^{''}} + x^{^{'}} + 10 \cdot x = 0

(2 D^{5} - 7 D^{4} + 12 D^{3} + 8 D^{2}) y = 0

y^{^{''}} + 22 y^{^{'}} + 2 y = 0

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

y^{^{''}} + 4 y^{^{'}} + 29 y = 0, y (0) = - 1, y^{^{'}} (0) = - 18

What is the solution for y^{''}-10y^'+50y=0 ?
The solution for y^{''}-10y^'+50y=0 is y=e^{5t}(c_{1}cos(5t)+c_{2}sin(5t))