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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{3}{4} e^{- x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{3}{4} e^{- x}

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{3}{4} e^{- x}

y^{^{''''}} + 9 y^{^{''}} = sec^{2} (3 t)

y^{^{''}} - 8 y^{^{'}} + 16 y = 8 x + 7

y^{^{''}} - 6 y^{^{'}} - 7 y = - 9 e^{- 2 x}

y^{^{''}} - 2 y^{^{'}} = - 2 e^{x}

y^{^{''}} - 16 y = (x + 1) e^{x}

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