What is the solution for x^{''}+2x=0,x(0)=-1,x^'(0)=-2sqrt(2) ?

The solution for x^{''}+2x=0,x(0)=-1,x^'(0)=-2sqrt(2) is x=-cos(sqrt(2)t)-2sin(sqrt(2)t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{''}+2x=0,x(0)=-1,x^'(0)=-2sqrt(2)

x^{^{''}} + 2 x = 0, x (0) = - 1, x^{^{'}} (0) = - 22

x = - cos (2 t) - 2 sin (2 t)

Solution steps

x^{''} (t) + 2 x = 0

Solve linear ODE:

x = - cos (2 t) - 2 sin (2 t)

x = - cos (2 t) - 2 sin (2 t)

y^{^{''}} = 3 y^{^{'}} + 5 y, y (2) = 8

y^{^{''}} + 36 y = 0, y (0) = 4, y^{^{'}} (0) = - 5

y^{^{''}} - y^{^{'}} - 72 y = 0, y (1) = e^{- 8 x}

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

y^{^{''}} + 10 y^{^{'}} - 24 y = 0

What is the solution for x^{''}+2x=0,x(0)=-1,x^'(0)=-2sqrt(2) ?
The solution for x^{''}+2x=0,x(0)=-1,x^'(0)=-2sqrt(2) is x=-cos(sqrt(2)t)-2sin(sqrt(2)t)