What is the solution for linear x^{''}+16x=0,x^'(0)=5 ?

The solution for linear x^{''}+16x=0,x^'(0)=5 is x=c_{1}cos(4t)+5/4 sin(4t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

linear x^{''}+16x=0,x^'(0)=5

linear

x^{^{''}} + 16 x = 0, x^{^{'}} (0) = 5

x = c_{1} cos (4 t) + \frac{5}{4} sin (4 t)

Solution steps

x^{''} (t) + 16 x = 0

Solve linear ODE:

x = c_{1} cos (4 t) + \frac{5}{4} sin (4 t)

x = c_{1} cos (4 t) + \frac{5}{4} sin (4 t)

y^{^{'''}} - 3 y^{^{''}} + 2 y^{^{'}} = 0

y^{^{''''}} - 6 y^{^{'''}} + 17 y^{^{''}} - 28 y^{^{'}} + 20 y = 0

y^{^{''}} + 9 y = 0, y (0) = 6, y^{^{'}} (0) = - 4

y^{^{''}} + 9 y = 0, y (0) = 6, y^{^{'}} (0) = - 5

y^{^{''}} - 3 y^{^{'}} - 5 y = 0, y (0) = - 2, y^{^{'}} (0) = 2

What is the solution for linear x^{''}+16x=0,x^'(0)=5 ?
The solution for linear x^{''}+16x=0,x^'(0)=5 is x=c_{1}cos(4t)+5/4 sin(4t)