What is the solution for y^'+4y=6e^{-3t} ?

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

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y^'+4y=6e^{-3t}

y^{^{'}} + 4 y = 6 e^{- 3 t}

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

Solution steps

y^{'} (t) + 4 y = 6 e^{- 3 t}

Solve linear ODE:

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

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

y \to - 3 lim (\frac{y ^{2} - 9}{2 y ^{2} + 7 y + 3})

16 y^{^{''}} - y = 0, y (- 7) = 1, y^{^{'}} (- 7) = - 1

\frac{d}{d x} (x^{\frac{5}{3}} - 5 x^{\frac{2}{3}})

\frac{d}{d x} (tan (x) + \frac{1}{x})

y = 8 x

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