What is the solution for y^'+4y=e^{-4t},y(0)=2 ?

The solution for y^'+4y=e^{-4t},y(0)=2 is y= t/(e^{4t)}+2/(e^{4t)}

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y^'+4y=e^{-4t},y(0)=2

y^{^{'}} + 4 y = e^{- 4 t}, y (0) = 2

y = \frac{t}{e ^{4 t}} + \frac{2}{e ^{4 t}}

Solution steps

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

Solve linear ODE:

y = \frac{t}{e ^{4 t}} + \frac{2}{e ^{4 t}}

y = \frac{t}{e ^{4 t}} + \frac{2}{e ^{4 t}}

y = \frac{3 + csc ( x )}{7 - csc ( x )}

\frac{2 x}{( x ^{2} - 5 ) ^{3}}, at x = 0

\int \frac{1}{x ^{3} x ^{2} - 81} d x

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

What is the solution for y^'+4y=e^{-4t},y(0)=2 ?
The solution for y^'+4y=e^{-4t},y(0)=2 is y= t/(e^{4t)}+2/(e^{4t)}