What is the solution for y^{''}-2y^'+y=(x2-2x)e^-x ?

The solution for y^{''}-2y^'+y=(x2-2x)e^-x is y=c_{1}e^x+c_{2}xe^x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-2y^'+y=(x2-2x)e^-x

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

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

Solution steps

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

Solve linear ODE:

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

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

- 3 y^{^{''}} + 27 y = 0, y^{^{'}} (- 2) = - 2, y (2) = - 2

y^{^{''}} + 5 y^{^{'}} = 0, y (0) = - 11, y^{^{'}} (0) = 35

12 y^{^{''}} - 11 y^{^{'}} + 2 y = 0, y (0) = 7, y^{^{'}} (0) = - 3

y^{^{''''}} - 4 y^{^{'''}} + 9 y^{^{''}} - 36 y^{^{'}} = 0

y^{^{''}} - 2 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

What is the solution for y^{''}-2y^'+y=(x2-2x)e^-x ?
The solution for y^{''}-2y^'+y=(x2-2x)e^-x is y=c_{1}e^x+c_{2}xe^x