What is the solution for y^{''}+4y=-cos(3x)-2sin(3x) ?

The solution for y^{''}+4y=-cos(3x)-2sin(3x) is y=c_{1}cos(2x)+c_{2}sin(2x)+2/5 sin(3x)+1/5 cos(3x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+4y=-cos(3x)-2sin(3x)

y^{^{''}} + 4 y = - cos (3 x) - 2 sin (3 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{2}{5} sin (3 x) + \frac{1}{5} cos (3 x)

Solution steps

y^{''} (x) + 4 y = - cos (3 x) - 2 sin (3 x)

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{2}{5} sin (3 x) + \frac{1}{5} cos (3 x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{2}{5} sin (3 x) + \frac{1}{5} cos (3 x)

What is the solution for y^{''}+4y=-cos(3x)-2sin(3x) ?
The solution for y^{''}+4y=-cos(3x)-2sin(3x) is y=c_{1}cos(2x)+c_{2}sin(2x)+2/5 sin(3x)+1/5 cos(3x)