What is the solution for y^'+3y=2e^{-3t} ?

The solution for y^'+3y=2e^{-3t} is y=(2t)/(e^{3t)}+(c_{1})/(e^{3t)}

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y^'+3y=2e^{-3t}

y^{^{'}} + 3 y = 2 e^{- 3 t}

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

Solution steps

y^{'} (t) + 3 y = 2 e^{- 3 t}

Solve linear ODE:

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

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

f (x) = \frac{( x - 2 )}{( 2 x ^{2} + 3 )}

\int_{2}^{3} \frac{1}{x ^{2} - 4 x} d x

\frac{d}{d x} (ln (1 - 6 x))

- x^{2} + 3 x + 4, at x = 1

What is the solution for y^'+3y=2e^{-3t} ?
The solution for y^'+3y=2e^{-3t} is y=(2t)/(e^{3t)}+(c_{1})/(e^{3t)}