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

The solution for y^{''}+4y^'+4y=x^2-2x is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(x^2)/4-x+7/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{x ^{2}}{4} - x + \frac{7}{8}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{x ^{2}}{4} - x + \frac{7}{8}

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{x ^{2}}{4} - x + \frac{7}{8}

y^{^{''}} + 25 y = 11 sec^{2} (5 t)

y^{^{''}} - 2 y^{^{'}} + 5 y = 3 e^{x} sin (x) + 24 x + 4

y^{^{'''}} - y^{^{'}} = 7 sin (t)

(D^{4} + D^{2}) y = 18 x - 4 sin (x)

y^{^{''}} - 2 y^{^{'}} + 5 y = e^{2 x}

What is the solution for y^{''}+4y^'+4y=x^2-2x ?
The solution for y^{''}+4y^'+4y=x^2-2x is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+(x^2)/4-x+7/8