What is the solution for x^2y^{''}-3xy^'+5y=0 ?

The solution for x^2y^{''}-3xy^'+5y=0 is y=x^2(c_{1}cos(ln(x))+c_{2}sin(ln(x)))

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x^2y^{''}-3xy^'+5y=0

x^{2} y^{^{''}} - 3 x y^{^{'}} + 5 y = 0

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

Solution steps

x^{2} y^{''} (x) - 3 x y^{'} (x) + 5 y = 0

Solve linear ODE:

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

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

f (x) = 2 x^{2} - 9 x + 10, at x = 2

\frac{d}{d x} (x arcsin (x))

tan (1 + e^{2 t})

\frac{\partial}{\partial x} (ln (5 x^{2} + 4 y^{2} + 9))

\int \frac{12 x ^{3}}{2 - x ^{4}} d x

What is the solution for x^2y^{''}-3xy^'+5y=0 ?
The solution for x^2y^{''}-3xy^'+5y=0 is y=x^2(c_{1}cos(ln(x))+c_{2}sin(ln(x)))