What is the solution for y^{''}+4y^'+4y=64cos(2x) ?

The solution for y^{''}+4y^'+4y=64cos(2x) is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+8sin(2x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'+4y=64cos(2x)

y^{^{''}} + 4 y^{^{'}} + 4 y = 64 cos (2 x)

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + 8 sin (2 x)

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = 64 cos (2 x)

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + 8 sin (2 x)

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + 8 sin (2 x)

\int (y - 2) d y

\frac{d}{d x} (\frac{cos ( 3 x )}{x})

y = (2 x - 1) (x + 4), (1, 5)

\frac{d}{d x} ((x^{3} - 2 x)^{x^{2} - 3})

What is the solution for y^{''}+4y^'+4y=64cos(2x) ?
The solution for y^{''}+4y^'+4y=64cos(2x) is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+8sin(2x)