What is the solution for y^{'''}+y^{''}-4=0 ?

The solution for y^{'''}+y^{''}-4=0 is y=c_{1}e^{-t}+c_{2}+c_{3}t+2t^2

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 6 y^{^{'}} + 8 y = 18 t e^{3 t}

y^{^{''}} - 8 y^{^{'}} - 4 y - 2 = 2 x

y^{^{''}} - 2 y^{^{''}} - 3 y = 4 e^{x} - 9

((d^{2}) - 1) y = x sin (x) + (1 + x^{2}) e^{x}

y^{^{''}} - 5 y^{^{'}} + 6 y = g (x)

What is the solution for y^{'''}+y^{''}-4=0 ?
The solution for y^{'''}+y^{''}-4=0 is y=c_{1}e^{-t}+c_{2}+c_{3}t+2t^2