What is the solution for x^3y^{'''}-6y=0 ?

The solution for x^3y^{'''}-6y=0 is y=c_{1}x^3+c_{2}cos(sqrt(2)ln(x))+c_{3}sin(sqrt(2)ln(x))

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x^3y^{'''}-6y=0

x^{3} y^{^{'''}} - 6 y = 0

y = c_{1} x^{3} + c_{2} cos (2 ln (x)) + c_{3} sin (2 ln (x))

Solution steps

x^{3} y^{'''} (x) - 6 y = 0

Solve linear ODE:

y = c_{1} x^{3} + c_{2} cos (2 ln (x)) + c_{3} sin (2 ln (x))

y = c_{1} x^{3} + c_{2} cos (2 ln (x)) + c_{3} sin (2 ln (x))

\int 7 arctan (3 y) d y

\int \frac{6 sin ( 2 x )}{sin ( x )} d x

n = 1 \sum \infty \frac{8}{π ^{n}}

\int sec^{2} (θ) + 7 d θ

y = \frac{5}{6} x^{2}, y = \frac{11}{6} - x^{2}

What is the solution for x^3y^{'''}-6y=0 ?
The solution for x^3y^{'''}-6y=0 is y=c_{1}x^3+c_{2}cos(sqrt(2)ln(x))+c_{3}sin(sqrt(2)ln(x))