What is the solution for y^{'''}=4t^3+t ?

The solution for y^{'''}=4t^3+t is y=(t^6}{30}+(t^4)/(24)+\frac{c_{1}t^2)/2+c_{2}t+c_{3}

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y^{'''}=4t^3+t

y^{^{'''}} = 4 t^{3} + t

y = \frac{t ^{6}}{30} + \frac{t ^{4}}{24} + \frac{c _{1} t ^{2}}{2} + c_{2} t + c_{3}

Solution steps

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

Solve

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

Solve

y^{''} (t) = t^{4} + \frac{t ^{2}}{2} + c_{1} : y^{'} (t) = \frac{t ^{5}}{5} + \frac{t ^{3}}{6} + c_{1} t + c_{2}

Solve

y^{'} (t) = \frac{t ^{5}}{5} + \frac{t ^{3}}{6} + c_{1} t + c_{2} : y = \frac{t ^{6}}{30} + \frac{t ^{4}}{24} + \frac{c _{1} t ^{2}}{2} + c_{2} t + c_{3}

y = \frac{t ^{6}}{30} + \frac{t ^{4}}{24} + \frac{c _{1} t ^{2}}{2} + c_{2} t + c_{3}

\frac{1}{1 + e ^{- t}}

\int yz e^{y} + 2 y d y

x \to 0 lim (\frac{1 - cos ( 3 x )}{x})

\frac{d y}{d x}, 9 e^{x y} + 4 x^{2} - 2 y^{2} = 36

ln (\frac{4 x + 1}{x ^{2} - 2})

What is the solution for y^{'''}=4t^3+t ?
The solution for y^{'''}=4t^3+t is y=(t^6}{30}+(t^4)/(24)+\frac{c_{1}t^2)/2+c_{2}t+c_{3}