What is the solution for y^'+4y=e^{-t}-4 ?

The solution for y^'+4y=e^{-t}-4 is y= 1/(3e^t)-1+(c_{1})/(e^{4t)}

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

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

y = \frac{1}{3 e ^{t}} - 1 + \frac{c _{1}}{e ^{4 t}}

Solution steps

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

Solve linear ODE:

y = \frac{1}{3 e ^{t}} - 1 + \frac{c _{1}}{e ^{4 t}}

y = \frac{1}{3 e ^{t}} - 1 + \frac{c _{1}}{e ^{4 t}}

\frac{d}{d x} (\frac{3}{2 3 x})

y = (5)^{x}, at x = 6

x \to - 2 lim (\frac{- 1}{( x + 2 )})

\frac{d}{d x} ((2 x - 1)^{3} (x + 1)^{5})

\frac{d}{d x} (\frac{1}{5})

What is the solution for y^'+4y=e^{-t}-4 ?
The solution for y^'+4y=e^{-t}-4 is y= 1/(3e^t)-1+(c_{1})/(e^{4t)}