What is the solution for 10y^{''}+60y^'+50y=5e^{-x} ?

The solution for 10y^{''}+60y^'+50y=5e^{-x} is y=c_{1}e^{-x}+c_{2}e^{-5x}+(e^{-x}x)/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

10y^{''}+60y^'+50y=5e^{-x}

10 y^{^{''}} + 60 y^{^{'}} + 50 y = 5 e^{- x}

y = c_{1} e^{- x} + c_{2} e^{- 5 x} + \frac{e ^{- x} x}{8}

Solution steps

10 y^{''} (x) + 60 y^{'} (x) + 50 y = 5 e^{- x}

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 5 x} + \frac{e ^{- x} x}{8}

y = c_{1} e^{- x} + c_{2} e^{- 5 x} + \frac{e ^{- x} x}{8}

y^{^{''}} - 16 y^{^{'}} + 64 y = \frac{2 e ^{8 x}}{3 x}

y^{^{''}} + 2 y^{^{'}} + 3 y = cos (2 t)

y^{^{''}} + 9 y = csc (3 x)

y^{^{''}} + y = cos (x), y (0) = 1, y^{^{'}} (0) = - 1

y^{^{''}} + 5 y^{^{'}} + 4 y = 8

What is the solution for 10y^{''}+60y^'+50y=5e^{-x} ?
The solution for 10y^{''}+60y^'+50y=5e^{-x} is y=c_{1}e^{-x}+c_{2}e^{-5x}+(e^{-x}x)/8