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

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

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

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

y = e^{34 t} (\frac{7 - 4 3 4 - 6 3 3 4 c _{3}}{18 3 4} + \frac{1}{3}) + e^{- \frac{3 4 t}{2}} - \frac{cos ( \frac{3 3 4 t}{2} ) ( 7 - 4 3 4 - 6 3 3 4 c _{3} )}{18 3 4} + c_{3} sin (\frac{3 3 4 t}{2}) - \frac{1}{3} e^{t} - \frac{t}{4}

Solution steps

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

Solve linear ODE:

y = e^{34 t} (\frac{7 - 4 3 4 - 6 3 3 4 c _{3}}{18 3 4} + \frac{1}{3}) + e^{- \frac{3 4 t}{2}} - \frac{cos ( \frac{3 3 4 t}{2} ) ( 7 - 4 3 4 - 6 3 3 4 c _{3} )}{18 3 4} + c_{3} sin (\frac{3 3 4 t}{2}) - \frac{1}{3} e^{t} - \frac{t}{4}

y = e^{34 t} (\frac{7 - 4 3 4 - 6 3 3 4 c _{3}}{18 3 4} + \frac{1}{3}) + e^{- \frac{3 4 t}{2}} - \frac{cos ( \frac{3 3 4 t}{2} ) ( 7 - 4 3 4 - 6 3 3 4 c _{3} )}{18 3 4} + c_{3} sin (\frac{3 3 4 t}{2}) - \frac{1}{3} e^{t} - \frac{t}{4}

y^{^{''}} - 2 y^{^{'}} - 3 y = 1, y (0) = 1, y^{^{'}} (0) = - 1

(D^{2} + 5 D + 4) Y = x^{2} + 7 x + 9

(d^{2} + 4) y = cos (x)

y^{^{''}} + 2 y^{^{'}} - 15 y = 7 e^{3 t} + (1 - 2 t) sin (6 t)

y^{^{''}} + 2 y^{^{'}} + 2 y = g (t), y (0) = 8, y^{^{'}} (0) = 6

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