What is the solution for y^'+5y=e^{-3t} ?

The solution for y^'+5y=e^{-3t} is y= 1/(2e^{3t)}+(c_{1})/(e^{5t)}

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y^'+5y=e^{-3t}

y^{^{'}} + 5 y = e^{- 3 t}

y = \frac{1}{2 e ^{3 t}} + \frac{c _{1}}{e ^{5 t}}

Solution steps

y^{'} (t) + 5 y = e^{- 3 t}

Solve linear ODE:

y = \frac{1}{2 e ^{3 t}} + \frac{c _{1}}{e ^{5 t}}

y = \frac{1}{2 e ^{3 t}} + \frac{c _{1}}{e ^{5 t}}

\frac{s + 2}{( s - 1 ) ( s + 3 ) ^{2}}

5 x - x^{2}

\int_{0}^{1} 23 (\frac{e ^{\frac{1}{x}}}{x ^{3}}) d x

x \to - 4 lim (\frac{1}{x + 4})

\int \frac{x + 1}{x + 10} d x

What is the solution for y^'+5y=e^{-3t} ?
The solution for y^'+5y=e^{-3t} is y= 1/(2e^{3t)}+(c_{1})/(e^{5t)}