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

The solution for x^3y^{'''}+6y=0 is y=(c_{1})/x+x^2(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 = \frac{c _{1}}{x} + x^{2} (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 = \frac{c _{1}}{x} + x^{2} (c_{2} cos (2 ln (x)) + c_{3} sin (2 ln (x)))

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

\int 4 sin (θ) d θ

x \to 1 lim (3 x^{3} - 4 x^{2} + x + 9)

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

f (x) = x^{2} + 5 x + 6

y^{2} y^{^{'}} + y^{3} = 1

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