What is the solution for y^{''}+4y^'+4y=4x+6 ?

The solution for y^{''}+4y^'+4y=4x+6 is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+x+1/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'+4y=4x+6

y^{^{''}} + 4 y^{^{'}} + 4 y = 4 x + 6

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

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = 4 x + 6

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} - 8 y = - 16 t + 16 t^{2}

y^{^{''}} - 16 y = 48 t - 24 e^{- 4 t}

x^{^{''}} + 2 x^{^{'}} + 50 x = 2 cos (7 t)

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

\frac{d ^{2} y}{d t ^{2}} + 4 y = 4 sec^{2} (2 t)

What is the solution for y^{''}+4y^'+4y=4x+6 ?
The solution for y^{''}+4y^'+4y=4x+6 is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+x+1/2