What is the solution for y^{''}+y-cos(2x)=0 ?

The solution for y^{''}+y-cos(2x)=0 is y=c_{1}cos(x)+c_{2}sin(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^{''}+y-cos(2x)=0

y^{^{''}} + y - cos (2 x) = 0

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

Solution steps

y^{''} (x) + y - cos (2 x) = 0

Solve linear ODE:

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

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

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