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

The solution for y^{'''}-y^'=e^{5t} is y=c_{1}+c_{2}e^{-t}+c_{3}e^t+1/120 e^{5t}

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

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

y = c_{1} + c_{2} e^{- t} + c_{3} e^{t} + \frac{1}{120} e^{5 t}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- t} + c_{3} e^{t} + \frac{1}{120} e^{5 t}

y = c_{1} + c_{2} e^{- t} + c_{3} e^{t} + \frac{1}{120} e^{5 t}

y^{^{''}} + 4 y^{^{'}} + 4 y = (x + 3) \cdot e^{(- 2 x)}

y^{^{''}} + 9 y = e^{2 t} H (t - 1)

y^{^{''}} + 2 y^{^{'}} + y = 1, y (0) = 2

y^{^{''}} - 3 y^{^{'}} + 2 y = x + 1 - 10 cos (3 x)

3 \frac{d ^{2} x}{d t ^{2}} + \frac{d x}{d t} + 10 x = 2 sin (3 t)

What is the solution for y^{'''}-y^'=e^{5t} ?
The solution for y^{'''}-y^'=e^{5t} is y=c_{1}+c_{2}e^{-t}+c_{3}e^t+1/120 e^{5t}