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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^'+2y=xe^{-2x}

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

y = \frac{x ^{2}}{2 e ^{2 x}} + \frac{c _{1}}{e ^{2 x}}

Solution steps

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

Solve linear ODE:

y = \frac{x ^{2}}{2 e ^{2 x}} + \frac{c _{1}}{e ^{2 x}}

y = \frac{x ^{2}}{2 e ^{2 x}} + \frac{c _{1}}{e ^{2 x}}

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