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

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

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(d^3y)/(dx^3)=4x^3+x

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

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

Solution steps

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

Substitute

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

with

y^{'''} (x)

y^{^{'''}} (x) = 4 x^{3} + x

Solve

y^{'''} (x) = 4 x^{3} + x : y^{''} (x) = x^{4} + \frac{x ^{2}}{2} + c_{1}

Solve

y^{''} (x) = x^{4} + \frac{x ^{2}}{2} + c_{1} : y^{'} (x) = \frac{x ^{5}}{5} + \frac{x ^{3}}{6} + c_{1} x + c_{2}

Solve

y^{'} (x) = \frac{x ^{5}}{5} + \frac{x ^{3}}{6} + c_{1} x + c_{2} : y = \frac{x ^{6}}{30} + \frac{x ^{4}}{24} + \frac{c _{1} x ^{2}}{2} + c_{2} x + c_{3}

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

\frac{d ^{2}}{d x ^{2}} (\frac{6}{7 x ^{4}})

\int 2 π x^{3} 1 + 9 x^{4} d x

\frac{\partial}{\partial l} (9 l^{\frac{2}{3}} k^{\frac{1}{3}})

y^{^{''}} + 9 y = 24 sin (3 t)

\int \frac{2 ^{x + 1}}{1 0 ^{x}} d x

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