What is the solution for y^{''}+3y^'+2y=5e^{-2x} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

y^{''} (x) + 3 y^{'} (x) + 2 y = 5 e^{- 2 x}

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - y = 12

y^{^{''}} + 36 y = 33 cos (5 t)

y^{^{''}} + 4 y = 5 (e^{x} + e^{3 x})

\frac{d ^{2}}{d x ^{2}} (θ (x)) - (a + b) θ (x) = (a \cdot c + b)

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

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