What is the solution for y^{''}+y=-cos(2t),y^'(0)=0,y(0)=0 ?

The solution for y^{''}+y=-cos(2t),y^'(0)=0,y(0)=0 is y=-1/3 cos(t)+1/3 cos(2t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+y=-cos(2t),y^'(0)=0,y(0)=0

y^{^{''}} + y = - cos (2 t), y^{^{'}} (0) = 0, y (0) = 0

y = - \frac{1}{3} cos (t) + \frac{1}{3} cos (2 t)

Solution steps

y^{''} (t) + y = - cos (2 t)

Solve linear ODE:

y = - \frac{1}{3} cos (t) + \frac{1}{3} cos (2 t)

y = - \frac{1}{3} cos (t) + \frac{1}{3} cos (2 t)

y^{^{''}} - 6 y^{^{'}} + 5 y = 12 e^{t}, y (0) = - 1, y^{^{'}} (0) = 7

y^{^{''}} + 9 y = 4 + 5 x

y^{^{''}} - 3 y^{^{'}} = 15

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = x^{- 4}

y^{^{'''}} - y^{^{'}} = t e^{- t}

What is the solution for y^{''}+y=-cos(2t),y^'(0)=0,y(0)=0 ?
The solution for y^{''}+y=-cos(2t),y^'(0)=0,y(0)=0 is y=-1/3 cos(t)+1/3 cos(2t)