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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = \frac{x}{e ^{2 x}} - \frac{1}{e ^{2 x}} + \frac{c _{1}}{e ^{3 x}}

Solution steps

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

Solve linear ODE:

y = \frac{x}{e ^{2 x}} - \frac{1}{e ^{2 x}} + \frac{c _{1}}{e ^{3 x}}

y = \frac{x}{e ^{2 x}} - \frac{1}{e ^{2 x}} + \frac{c _{1}}{e ^{3 x}}

\frac{\partial}{\partial y} (- sin (x) sin (y))

n = 0 \sum \infty \frac{( x ^{n} )}{n}

\frac{d y}{y ln ( \frac{k}{y} )} = c d x

(2 x + 1)^{10}

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