What is the solution for y^{''}+4y=cos(x),y(0)=0,y^'(0)=0 ?

The solution for y^{''}+4y=cos(x),y(0)=0,y^'(0)=0 is y= 1/3 cos(x)-1/3 cos(2x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+4y=cos(x),y(0)=0,y^'(0)=0

y^{^{''}} + 4 y = cos (x), y (0) = 0, y^{^{'}} (0) = 0

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

Solution steps

y^{''} (x) + 4 y = cos (x)

Solve linear ODE:

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

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

What is the solution for y^{''}+4y=cos(x),y(0)=0,y^'(0)=0 ?
The solution for y^{''}+4y=cos(x),y(0)=0,y^'(0)=0 is y= 1/3 cos(x)-1/3 cos(2x)