What is the solution for y^{''''}+1=0 ?

The solution for y^{''''}+1=0 is y=-(t^4}{24}+\frac{c_{1}t^3)/6+(c_{2}t^2)/2+c_{3}t+c_{4}

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y^{''''}+1=0

y^{^{''''}} + 1 = 0

y = - \frac{t ^{4}}{24} + \frac{c _{1} t ^{3}}{6} + \frac{c _{2} t ^{2}}{2} + c_{3} t + c_{4}

Solution steps

y^{(4)} (t) + 1 = 0

Isolate

y^{(4)} (t) : y^{(4)} (t) = - 1

Solve

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

Solve

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

Solve

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

Solve

y^{'} (t) = - \frac{t ^{3}}{6} + \frac{c _{1} t ^{2}}{2} + c_{2} t + c_{3} : y = - \frac{t ^{4}}{24} + \frac{c _{1} t ^{3}}{6} + \frac{c _{2} t ^{2}}{2} + c_{3} t + c_{4}

y = - \frac{t ^{4}}{24} + \frac{c _{1} t ^{3}}{6} + \frac{c _{2} t ^{2}}{2} + c_{3} t + c_{4}

\frac{\partial}{\partial x} (\frac{4}{x y} - x^{2} + 3 y^{2})

\frac{d}{d x} (\frac{1 + x ^{2}}{( 1 - x ^{2} ) ^{2}})

5 x^{\frac{7}{4}} + 5 x^{\frac{1}{3}} - 2

(2 x + t) \frac{d x}{d t} + x - 2 t = 0

\int e^{- \frac{x}{b}} d x

What is the solution for y^{''''}+1=0 ?
The solution for y^{''''}+1=0 is y=-(t^4}{24}+\frac{c_{1}t^3)/6+(c_{2}t^2)/2+c_{3}t+c_{4}