What is the solution for y^{'''}-8=0 ?

The solution for y^{'''}-8=0 is y=(4t^3)/3+(c_{1}t^2)/2+c_{2}t+c_{3}

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y^{'''}-8=0

y^{^{'''}} - 8 = 0

y = \frac{4 t ^{3}}{3} + \frac{c _{1} t ^{2}}{2} + c_{2} t + c_{3}

Solution steps

y^{'''} (t) - 8 = 0

Isolate

y^{'''} (t) : y^{'''} (t) = 8

Solve

y^{'''} (t) = 8 : y^{''} (t) = 8 t + c_{1}

Solve

y^{''} (t) = 8 t + c_{1} : y^{'} (t) = 4 t^{2} + c_{1} t + c_{2}

Solve

y^{'} (t) = 4 t^{2} + c_{1} t + c_{2} : y = \frac{4 t ^{3}}{3} + \frac{c _{1} t ^{2}}{2} + c_{2} t + c_{3}

y = \frac{4 t ^{3}}{3} + \frac{c _{1} t ^{2}}{2} + c_{2} t + c_{3}

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f (x) = 5 x^{2} - 5 x, at x = 3

\int \frac{arctan ( \frac{x}{2} )}{4 + x ^{2}} d x

What is the solution for y^{'''}-8=0 ?
The solution for y^{'''}-8=0 is y=(4t^3)/3+(c_{1}t^2)/2+c_{2}t+c_{3}