What is the solution for y^'+3y=e^{-4t},y(0)=5 ?

The solution for y^'+3y=e^{-4t},y(0)=5 is y=-1/(e^{4t)}+6/(e^{3t)}

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y^'+3y=e^{-4t},y(0)=5

y^{^{'}} + 3 y = e^{- 4 t}, y (0) = 5

y = - \frac{1}{e ^{4 t}} + \frac{6}{e ^{3 t}}

Solution steps

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

Solve linear ODE:

y = - \frac{1}{e ^{4 t}} + \frac{6}{e ^{3 t}}

y = - \frac{1}{e ^{4 t}} + \frac{6}{e ^{3 t}}

\frac{d}{d x} (\frac{9}{e ^{x} - e ^{- x}})

x^{2} + 3 y \frac{d y}{d x} = 0

\int_{0}^{16} \frac{1}{x} d x

y^{^{'}} = ry

y^{^{''}} = - \frac{2}{t ^{2}}

What is the solution for y^'+3y=e^{-4t},y(0)=5 ?
The solution for y^'+3y=e^{-4t},y(0)=5 is y=-1/(e^{4t)}+6/(e^{3t)}