What is the solution for x^2y^{''}-2y=sin(ln(x)) ?

The solution for x^2y^{''}-2y=sin(ln(x)) is y=c_{1}x^2+(c_{2})/x+(-3sin(ln(x))+cos(ln(x)))/(10)

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x^2y^{''}-2y=sin(ln(x))

x^{2} y^{^{''}} - 2 y = sin (ln (x))

y = c_{1} x^{2} + \frac{c _{2}}{x} + \frac{- 3 sin ( ln ( x ) ) + cos ( ln ( x ) )}{10}

Solution steps

x^{2} y^{''} (x) - 2 y = sin (ln (x))

Solve linear ODE:

y = c_{1} x^{2} + \frac{c _{2}}{x} + \frac{- 3 sin ( ln ( x ) ) + cos ( ln ( x ) )}{10}

y = c_{1} x^{2} + \frac{c _{2}}{x} + \frac{- 3 sin ( ln ( x ) ) + cos ( ln ( x ) )}{10}

x \to - 2 lim (2 x)

\int \frac{1}{5 x ^{7}} d x

ma c l a u r in \frac{arctan ( x )}{x + 1}

\frac{\partial}{\partial y} (y - cos (y))

d er i v a t i v e f (x) = (x + 2)^{x}

What is the solution for x^2y^{''}-2y=sin(ln(x)) ?
The solution for x^2y^{''}-2y=sin(ln(x)) is y=c_{1}x^2+(c_{2})/x+(-3sin(ln(x))+cos(ln(x)))/(10)