What is the solution for y^'+3y=e^{-4t} ?

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

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y^'+3y=e^{-4t}

y^{^{'}} + 3 y = e^{- 4 t}

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

Solution steps

y^{'} (t) + 3 y = e^{- 4 t}

Solve linear ODE:

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

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

\int_{0}^{2} 4 x - x^{3} d x

e^{t} (cos (t))

x = - 2, x = 2, y = 2 x^{2} + 5, y = 0

sin (- 1 (6 x + 1))

\frac{2}{s ( s + 2 )}

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