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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int_{0}^{1} 3 x^{3} d x

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

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

\int sin^{4} (x co) s^{5} x d x

h \to 0 lim (\frac{( 1 + h ) ^{5} - 1}{h})

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