What is the solution for y^{''}+3y^'+2y=4x ?

The solution for y^{''}+3y^'+2y=4x is y=c_{1}e^{-x}+c_{2}e^{-2x}+2x-3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + 2 x - 3

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + 2 x - 3

y = c_{1} e^{- x} + c_{2} e^{- 2 x} + 2 x - 3

(D^{2} - 4 D + 4) y = e^{x}

y^{^{''}} - 3 y^{^{'}} + 2 y = 12 e^{- 2 t}

3 y^{^{''}} + 12 y^{^{'}} + 39 y = 90 sin (3 t)

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

y^{^{''}} - 2 y^{^{'}} + 6 = 0

What is the solution for y^{''}+3y^'+2y=4x ?
The solution for y^{''}+3y^'+2y=4x is y=c_{1}e^{-x}+c_{2}e^{-2x}+2x-3