What is the solution for y^{''}-8y^'-20y=270e^x ?

The solution for y^{''}-8y^'-20y=270e^x is y=c_{1}e^{10x}+c_{2}e^{-2x}-10e^x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-8y^'-20y=270e^x

y^{^{''}} - 8 y^{^{'}} - 20 y = 270 e^{x}

y = c_{1} e^{10 x} + c_{2} e^{- 2 x} - 10 e^{x}

Solution steps

y^{''} (x) - 8 y^{'} (x) - 20 y = 270 e^{x}

Solve linear ODE:

y = c_{1} e^{10 x} + c_{2} e^{- 2 x} - 10 e^{x}

y = c_{1} e^{10 x} + c_{2} e^{- 2 x} - 10 e^{x}

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

y^{^{''}} + 10 y^{^{'}} + 25 y = x, y (0) = 0, y^{^{'}} (0) = 0

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

3 y^{^{''}} + 27 y = 2 cos (3 x)

(3 D^{2} + 3) y = 3 x

What is the solution for y^{''}-8y^'-20y=270e^x ?
The solution for y^{''}-8y^'-20y=270e^x is y=c_{1}e^{10x}+c_{2}e^{-2x}-10e^x