What is the solution for (d^3x)/(dt^3)=4t^3+t ?

The solution for (d^3x)/(dt^3)=4t^3+t is x=(t^6}{30}+(t^4)/(24)+\frac{c_{1}t^2)/2+c_{2}t+c_{3}

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(d^3x)/(dt^3)=4t^3+t

\frac{d ^{3} x}{d t ^{3}} = 4 t^{3} + t

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

Solution steps

\frac{d ^{3} x}{d t ^{3}} = 4 t^{3} + t

Substitute

\frac{d ^{3} x}{d t ^{3}}

with

x^{'''} (t)

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

Solve

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

Solve

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

Solve

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

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

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

\int 5 \cdot sin (x) d x

(x^{2} + 1) \frac{d y}{d x} + 8 x (y - 1) = 0, y (0) = 2

\int \frac{1}{8 - x ^{2}} d x

x^{2} - 4 x + 6

What is the solution for (d^3x)/(dt^3)=4t^3+t ?
The solution for (d^3x)/(dt^3)=4t^3+t is x=(t^6}{30}+(t^4)/(24)+\frac{c_{1}t^2)/2+c_{2}t+c_{3}