What is the solution for y^{'''}+y^{''}=1 ?

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

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y^{'''}+y^{''}=1

y^{^{'''}} + y^{^{''}} = 1

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

Solution steps

y^{'''} (t) + y^{''} (t) = 1

Solve linear ODE:

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

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

(d^{3} + d) \cdot y = cos (x)

(d^{3} + 3 d^{2} - 4) y = sinh^{2} (x) + 7

y^{^{''}} + 4 y^{^{'}} = 12 x^{2} - 10 x - 16

(d^{2} + 6 d + 9) y = 5 e^{3} x

(d^{3} - 2 d^{2} - 3 d + 10) y = 40 cos (x)

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