What is the solution for y^{'''}+16y^'=t ?

The solution for y^{'''}+16y^'=t is y=c_{1}+c_{2}cos(4t)+c_{3}sin(4t)+(t^2)/(32)

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y^{'''}+16y^'=t

y^{^{'''}} + 16 y^{^{'}} = t

y = c_{1} + c_{2} cos (4 t) + c_{3} sin (4 t) + \frac{t ^{2}}{32}

Solution steps

y^{'''} (t) + 16 y^{'} (t) = t

Solve linear ODE:

y = c_{1} + c_{2} cos (4 t) + c_{3} sin (4 t) + \frac{t ^{2}}{32}

y = c_{1} + c_{2} cos (4 t) + c_{3} sin (4 t) + \frac{t ^{2}}{32}

y^{^{''}} + 3 y^{^{'}} + 2 y = \frac{x}{e ^{x}} + \frac{x ^{2}}{e ^{2 x}}

y^{^{''}} + 2 y^{^{'}} + y = 12 e^{- x}

y^{^{''''}} - 5 y^{^{''}} + 4 = 0

\frac{y ^{^{''}}}{8} + 32 y = 2 cos (3 t) + 4

y^{^{''}} - 3 y^{^{'}} + 2 y = x, y (0) = e^{x}

What is the solution for y^{'''}+16y^'=t ?
The solution for y^{'''}+16y^'=t is y=c_{1}+c_{2}cos(4t)+c_{3}sin(4t)+(t^2)/(32)