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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} - y^{^{''}} - 16 y^{^{'}} + 16 y = 8 - e^{x} + e^{4 x}

y^{^{''}} - 2 y^{^{'}} + y = - 4 e^{x} + 3 x^{2}

y^{^{''}} + 3 y^{^{'}} + 4 y = 8 x + 2

(d^{2} + d) y = x^{2} + 2 x - 3

2 y^{^{''}} + 12 y = 6 tan (3 x)

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