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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + y = cos (5 t)

y^{^{''}} + 5 y^{^{'}} = 2 x^{3} - 4 x^{2} - x + 6

y^{^{''}} + 8 y^{^{'}} + 16 y = 3 \frac{e ^{- 4 x}}{x ^{4}}

y^{^{''}} + 2 \cdot y^{^{'}} + y = e^{- t}

y^{^{''}} - y = 6 e^{2 x}, y (0) = 4, y^{^{'}} (0) = 4

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