What is the solution for y^{''}+6y^'+9y=(e^{-3}x)/x ?

The solution for y^{''}+6y^'+9y=(e^{-3}x)/x is y=c_{1}e^{-3x}+c_{2}xe^{-3x}+1/(9e^3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+6y^'+9y=(e^{-3}x)/x

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{e ^{- 3} x}{x}

y = c_{1} e^{- 3 x} + c_{2} x e^{- 3 x} + \frac{1}{9 e ^{3}}

Solution steps

y^{''} (x) + 6 y^{'} (x) + 9 y = \frac{e ^{- 3} x}{x}

Solve linear ODE:

y = c_{1} e^{- 3 x} + c_{2} x e^{- 3 x} + \frac{1}{9 e ^{3}}

y = c_{1} e^{- 3 x} + c_{2} x e^{- 3 x} + \frac{1}{9 e ^{3}}

(d^{3} + d^{2} - 4 d - 4) y = 3 e^{- x} - 4 x - 6

(d^{2} - 2 d + 4) \cdot y = e^{x} \cdot cos (x)

y^{^{'''}} - 3 y^{^{''}} - 10 y^{^{'}} + 24 = 2 x^{2} - 5

(d^{2} - 3 d + 2) y = sin (2 x)

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

What is the solution for y^{''}+6y^'+9y=(e^{-3}x)/x ?
The solution for y^{''}+6y^'+9y=(e^{-3}x)/x is y=c_{1}e^{-3x}+c_{2}xe^{-3x}+1/(9e^3)