What is the solution for y^{''}+4y^'+4y=(e^{-2x})/(x^3) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = \frac{e ^{- 2 x}}{x ^{3}}

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} - 6 y = - 22 t + 12 t^{2}

y^{^{''}} + 4 y = - t^{2}

x^{^{''}} - 9 x = - 13 cos (2 t)

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

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

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