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

The solution for y^{''''}=0 is y=(c_{1}t^3)/6+(c_{2}t^2)/2+c_{3}t+c_{4}

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y^{''''}=0

y^{^{''''}} = 0

y = \frac{c _{1} t ^{3}}{6} + \frac{c _{2} t ^{2}}{2} + c_{3} t + c_{4}

Solution steps

y^{(4)} (t) = 0

Solve

y^{(4)} (t) = 0 : y^{'''} (t) = c_{1}

Solve

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

Solve

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

Solve

y^{'} (t) = \frac{c _{1} t ^{2}}{2} + c_{2} t + c_{3} : y = \frac{c _{1} t ^{3}}{6} + \frac{c _{2} t ^{2}}{2} + c_{3} t + c_{4}

y = \frac{c _{1} t ^{3}}{6} + \frac{c _{2} t ^{2}}{2} + c_{3} t + c_{4}

y^{^{''}} - 6 y^{^{'}} + 25 y = 64 e^{- t}

\int 2 (x + 2)^{3} d x

\frac{\partial}{\partial x} (x^{3} + 3 x y^{2} - 15 x - 12 y)

\frac{d ^{2} y}{d t ^{2}} - 5 \frac{d y}{d t} - 24 y = 0

x \to 0 - lim (x - 1)

What is the solution for y^{''''}=0 ?
The solution for y^{''''}=0 is y=(c_{1}t^3)/6+(c_{2}t^2)/2+c_{3}t+c_{4}