What is the solution for y^{'''}-2y^{''}=4 ?

The solution for y^{'''}-2y^{''}=4 is y=c_{1}e^{2t}+c_{2}+c_{3}t-t^2

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y^{'''}-2y^{''}=4

y^{^{'''}} - 2 y^{^{''}} = 4

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

Solution steps

y^{'''} (t) - 2 y^{''} (t) = 4

Solve linear ODE:

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

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

y^{^{''}} - 9 y = e^{(- 2 x)} (- 17 + 10 x)

y^{^{''}} + 2 y^{^{'}} - 3 y = x^{2} sin (x)

y^{^{''}} + 5 y = - 9.81

y^{^{''}} + 4 y^{^{'}} + 4 y = 3 e^{- 3 x}

\frac{d ^{2} y}{d x ^{2}} + 5 \frac{d y}{d x} + 6 y = x e^{3 x}

What is the solution for y^{'''}-2y^{''}=4 ?
The solution for y^{'''}-2y^{''}=4 is y=c_{1}e^{2t}+c_{2}+c_{3}t-t^2