What is the solution for x^2y^{''}-3xy^'+13y=0 ?

The solution for x^2y^{''}-3xy^'+13y=0 is y=x^2(c_{1}cos(3ln(x))+c_{2}sin(3ln(x)))

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x^2y^{''}-3xy^'+13y=0

x^{2} y^{^{''}} - 3 x y^{^{'}} + 13 y = 0

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

Solution steps

x^{2} y^{''} (x) - 3 x y^{'} (x) + 13 y = 0

Solve linear ODE:

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

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

\int arccot (9 y) d y

\int \frac{12 x ^{5}}{3 + 2 x ^{6}} d x

\frac{d}{d x} (sin^{4} (2 x))

\frac{d}{d x} (\frac{( x + 1 ) ( 2 x - 7 )}{2 x + 1})

b \to \infty lim (e^{b})

What is the solution for x^2y^{''}-3xy^'+13y=0 ?
The solution for x^2y^{''}-3xy^'+13y=0 is y=x^2(c_{1}cos(3ln(x))+c_{2}sin(3ln(x)))