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)}

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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}}

\frac{d}{d x} (e^{y} sin (x y))

x \to - \infty lim (2 x - x^{2})

\int 2 x x^{2} - 1 d x

n = 0 \sum \infty (- 0.6)^{n}

\frac{0.3 ( 5 ^{x} )}{x ^{3}}

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)}