What is the solution for y^'+3y=2e^{-3t},y(0)=2 ?

The solution for y^'+3y=2e^{-3t},y(0)=2 is y=(2t)/(e^{3t)}+2/(e^{3t)}

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y^'+3y=2e^{-3t},y(0)=2

y^{^{'}} + 3 y = 2 e^{- 3 t}, y (0) = 2

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

Solution steps

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

Solve linear ODE:

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

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

\int 3 x (x - 4) d x

\frac{d}{d x} (x^{2} z (x))

x \to e lim (x^{2} \cdot e^{x^{2}})

f (x) = 2 e^{x} \cdot (x^{2} + 3 x - 9)^{2}

\frac{d}{d θ} (\frac{cos ( θ )}{sin ( θ )})

What is the solution for y^'+3y=2e^{-3t},y(0)=2 ?
The solution for y^'+3y=2e^{-3t},y(0)=2 is y=(2t)/(e^{3t)}+2/(e^{3t)}