What is the solution for y^{''}+10y^'+24y=e^x+3x ?

The solution for y^{''}+10y^'+24y=e^x+3x is y=c_{1}e^{-4x}+c_{2}e^{-6x}+1/35 e^x+x/8-5/96

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+10y^'+24y=e^x+3x

y^{^{''}} + 10 y^{^{'}} + 24 y = e^{x} + 3 x

y = c_{1} e^{- 4 x} + c_{2} e^{- 6 x} + \frac{1}{35} e^{x} + \frac{x}{8} - \frac{5}{96}

Solution steps

y^{''} (x) + 10 y^{'} (x) + 24 y = e^{x} + 3 x

Solve linear ODE:

y = c_{1} e^{- 4 x} + c_{2} e^{- 6 x} + \frac{1}{35} e^{x} + \frac{x}{8} - \frac{5}{96}

y = c_{1} e^{- 4 x} + c_{2} e^{- 6 x} + \frac{1}{35} e^{x} + \frac{x}{8} - \frac{5}{96}

y^{^{''}} + k^{2} y = sin (b x)

y^{^{''}} + y = cosh (x)

(D^{2} + D - 2) y = 2 x - 40 cos (2 x)

y^{^{''''}} + 16 y^{^{''}} = sin (4 t) + t e^{5 t} + 4

y^{^{''}} - 4 y^{^{'}} + 3 y = e^{3 x}

What is the solution for y^{''}+10y^'+24y=e^x+3x ?
The solution for y^{''}+10y^'+24y=e^x+3x is y=c_{1}e^{-4x}+c_{2}e^{-6x}+1/35 e^x+x/8-5/96