What is the solution for y^{'''}=x^4 ?

The solution for y^{'''}=x^4 is y=(x^7}{210}+\frac{c_{1}x^2)/2+c_{2}x+c_{3}

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y^{'''}=x^4

y^{^{'''}} = x^{4}

y = \frac{x ^{7}}{210} + \frac{c _{1} x ^{2}}{2} + c_{2} x + c_{3}

Solution steps

y^{'''} (x) = x^{4}

Solve

y^{'''} (x) = x^{4} : y^{''} (x) = \frac{x ^{5}}{5} + c_{1}

Solve

y^{''} (x) = \frac{x ^{5}}{5} + c_{1} : y^{'} (x) = \frac{x ^{6}}{30} + c_{1} x + c_{2}

Solve

y^{'} (x) = \frac{x ^{6}}{30} + c_{1} x + c_{2} : y = \frac{x ^{7}}{210} + \frac{c _{1} x ^{2}}{2} + c_{2} x + c_{3}

y = \frac{x ^{7}}{210} + \frac{c _{1} x ^{2}}{2} + c_{2} x + c_{3}

\frac{y ^{^{''}} + 1}{y ^{2}} = 0

y^{^{'}} - e^{4} x = 0

x \to 3 lim ((\frac{1 + 1}{x})^{x})

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

x \to 3 lim (\frac{1}{cot ^{3} ( x )})

What is the solution for y^{'''}=x^4 ?
The solution for y^{'''}=x^4 is y=(x^7}{210}+\frac{c_{1}x^2)/2+c_{2}x+c_{3}