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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+y=cos(2t),y(0)=0,y^'(0)=1

y^{^{''}} + y = cos (2 t), y (0) = 0, y^{^{'}} (0) = 1

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

Solution steps

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

Solve linear ODE:

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

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

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