What is the solution for y^{''}-2y^'-8y=(x+1)e^x ?

The solution for y^{''}-2y^'-8y=(x+1)e^x is y=c_{1}e^{4x}+c_{2}e^{-2x}-(e^xx)/9-1/9 e^x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-2y^'-8y=(x+1)e^x

y^{^{''}} - 2 y^{^{'}} - 8 y = (x + 1) e^{x}

y = c_{1} e^{4 x} + c_{2} e^{- 2 x} - \frac{e ^{x} x}{9} - \frac{1}{9} e^{x}

Solution steps

y^{''} (x) - 2 y^{'} (x) - 8 y = (x + 1) e^{x}

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{- 2 x} - \frac{e ^{x} x}{9} - \frac{1}{9} e^{x}

y = c_{1} e^{4 x} + c_{2} e^{- 2 x} - \frac{e ^{x} x}{9} - \frac{1}{9} e^{x}

y^{^{''}} - y^{^{'}} - 6 y = 56 cos (3 x)

(d^{2} + 6 d + 10) y = 3 e^{- 3 x} x - 2 e^{3 x} cos (x)

D u = 50 - \frac{30}{100} \frac{1}{500}

(d^{2} - 1) y = e^{- x} (2 sin (x) + 4 cos (x))

y^{^{''}} + 4 y^{^{'}} + 4 y = \frac{6 e ^{- 2 x}}{x ^{5}}

What is the solution for y^{''}-2y^'-8y=(x+1)e^x ?
The solution for y^{''}-2y^'-8y=(x+1)e^x is y=c_{1}e^{4x}+c_{2}e^{-2x}-(e^xx)/9-1/9 e^x