What is the solution for y^{''}+4y^'+3y=4e^{-x} ?

The solution for y^{''}+4y^'+3y=4e^{-x} is y=c_{1}e^{-x}+c_{2}e^{-3x}+2xe^{-x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (x \cdot sin (x + y))

\int (1 - sin (x)) d x

y = 6 tan (x), at x = \frac{π}{3}

\frac{d}{d x} (\frac{tan ( \frac{x}{2} ) - 3}{3})

y = es (s^{3} + 10), (0, 10)

What is the solution for y^{''}+4y^'+3y=4e^{-x} ?
The solution for y^{''}+4y^'+3y=4e^{-x} is y=c_{1}e^{-x}+c_{2}e^{-3x}+2xe^{-x}