What is the solution for y^{'''}+3y^{''}-4=0 ?

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

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y^{'''}+3y^{''}-4=0

y^{^{'''}} + 3 y^{^{''}} - 4 = 0

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

Solution steps

y^{'''} (t) + 3 y^{''} (t) - 4 = 0

Solve linear ODE:

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

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

2 y^{^{''}} + 8 y^{^{'}} - y = 8

y^{^{''}} + 3 y^{^{'}} + 2 y = 3 e^{- 4 t}, y (0) = 1, y^{^{'}} (0) = 0

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

y^{^{''}} + y^{^{'}} - 2 y = 9 e^{x} - 2

\frac{d ^{2} x}{d t ^{2}} + 3 \frac{d x}{d t} + 2 x = 2 e^{- 3 t}

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