What is the solution for y^{'''}-2y^{''}+y^'=3t^2-18 ?

The solution for y^{'''}-2y^{''}+y^'=3t^2-18 is y=c_{1}+c_{2}e^t+c_{3}te^t+t^3+6t^2

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y^{'''}-2y^{''}+y^'=3t^2-18

y^{^{'''}} - 2 y^{^{''}} + y^{^{'}} = 3 t^{2} - 18

y = c_{1} + c_{2} e^{t} + c_{3} t e^{t} + t^{3} + 6 t^{2}

Solution steps

y^{'''} (t) - 2 y^{''} (t) + y^{'} (t) = 3 t^{2} - 18

Solve linear ODE:

y = c_{1} + c_{2} e^{t} + c_{3} t e^{t} + t^{3} + 6 t^{2}

y = c_{1} + c_{2} e^{t} + c_{3} t e^{t} + t^{3} + 6 t^{2}

y^{^{''}} + 8 y^{^{'}} + 2 y = cos (x), y (0) = 0

y^{^{''}} + y^{^{'}} + y 2 = sin (x^{2})

10 x^{^{''}} + 45 x^{^{'}} + 2500 x = 180 cos (7 π)

y^{^{''}} + 7 y^{^{'}} + 12 y = 5 x

\frac{d ^{2} y}{d x ^{2}} - 16 \frac{d y}{d x} - 64 y = 7 e^{8 x}

What is the solution for y^{'''}-2y^{''}+y^'=3t^2-18 ?
The solution for y^{'''}-2y^{''}+y^'=3t^2-18 is y=c_{1}+c_{2}e^t+c_{3}te^t+t^3+6t^2