What is the solution for y^{''}+8y^'+15y=2e^{-5x} ?

The solution for y^{''}+8y^'+15y=2e^{-5x} is y=c_{1}e^{-3x}+c_{2}e^{-5x}-e^{-5x}x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+8y^'+15y=2e^{-5x}

y^{^{''}} + 8 y^{^{'}} + 15 y = 2 e^{- 5 x}

y = c_{1} e^{- 3 x} + c_{2} e^{- 5 x} - e^{- 5 x} x

Solution steps

y^{''} (x) + 8 y^{'} (x) + 15 y = 2 e^{- 5 x}

Solve linear ODE:

y = c_{1} e^{- 3 x} + c_{2} e^{- 5 x} - e^{- 5 x} x

y = c_{1} e^{- 3 x} + c_{2} e^{- 5 x} - e^{- 5 x} x

\frac{d}{d x} (\frac{1}{x ^{2} + y ^{2} - 25})

\int e^{t x - x} d x

\frac{\partial}{\partial y} (7 x + 8 y^{2})

\int e^{x} cos (y) + yz d x

\frac{d}{d x} (x!)

What is the solution for y^{''}+8y^'+15y=2e^{-5x} ?
The solution for y^{''}+8y^'+15y=2e^{-5x} is y=c_{1}e^{-3x}+c_{2}e^{-5x}-e^{-5x}x