What is the solution for x^'+4x=te^{-4t} ?

The solution for x^'+4x=te^{-4t} is x=(t^2)/(2e^{4t)}+(c_{1})/(e^{4t)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

x^'+4x=te^{-4t}

x^{^{'}} + 4 x = t e^{- 4 t}

x = \frac{t ^{2}}{2 e ^{4 t}} + \frac{c _{1}}{e ^{4 t}}

Solution steps

x^{'} (t) + 4 x = t e^{- 4 t}

Solve linear ODE:

x = \frac{t ^{2}}{2 e ^{4 t}} + \frac{c _{1}}{e ^{4 t}}

x = \frac{t ^{2}}{2 e ^{4 t}} + \frac{c _{1}}{e ^{4 t}}

What is the solution for x^'+4x=te^{-4t} ?
The solution for x^'+4x=te^{-4t} is x=(t^2)/(2e^{4t)}+(c_{1})/(e^{4t)}