What is the solution for y^{'''}+y^{''}-2=0 ?

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

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y^{'''}+y^{''}-2=0

y^{^{'''}} + y^{^{''}} - 2 = 0

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

Solution steps

y^{'''} (t) + y^{''} (t) - 2 = 0

Solve linear ODE:

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

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

\int \frac{x ^{2} + 2 x + 2}{x + 1} d x

\frac{d}{d x} (sec^{2} (\frac{x}{a}) + csc^{2} (\frac{x}{a}))

f (x) = x^{2} - 6, (2, - 2)

x \to - \infty lim (2. 3^{x})

2 y^{^{''}} + 35 y = 0

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