What is the solution for y^{''}+2y^'+5y=52cos(3x) ?

The solution for y^{''}+2y^'+5y=52cos(3x) is y=e^{-x}(c_{1}cos(2x)+c_{2}sin(2x))+6sin(3x)-4cos(3x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+2y^'+5y=52cos(3x)

y^{^{''}} + 2 y^{^{'}} + 5 y = 52 cos (3 x)

y = e^{- x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + 6 sin (3 x) - 4 cos (3 x)

Solution steps

y^{''} (x) + 2 y^{'} (x) + 5 y = 52 cos (3 x)

Solve linear ODE:

y = e^{- x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + 6 sin (3 x) - 4 cos (3 x)

y = e^{- x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + 6 sin (3 x) - 4 cos (3 x)

What is the solution for y^{''}+2y^'+5y=52cos(3x) ?
The solution for y^{''}+2y^'+5y=52cos(3x) is y=e^{-x}(c_{1}cos(2x)+c_{2}sin(2x))+6sin(3x)-4cos(3x)