What is the solution for x^2y^{''}+xy^'-y= 1/x ?

The solution for x^2y^{''}+xy^'-y= 1/x is y=c_{1}x+(c_{2})/x-1/(4x)-(ln(x))/(2x)

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x^2y^{''}+xy^'-y= 1/x

x^{2} y^{^{''}} + x y^{^{'}} - y = \frac{1}{x}

y = c_{1} x + \frac{c _{2}}{x} - \frac{1}{4 x} - \frac{ln ( x )}{2 x}

Solution steps

x^{2} y^{''} (x) + x y^{'} (x) - y = \frac{1}{x}

Solve linear ODE:

y = c_{1} x + \frac{c _{2}}{x} - \frac{1}{4 x} - \frac{ln ( x )}{2 x}

y = c_{1} x + \frac{c _{2}}{x} - \frac{1}{4 x} - \frac{ln ( x )}{2 x}

\frac{d y}{d x} = (\frac{4 y + 3}{8 x + 5})^{2}

\int cos (x^{2}) x d x

\frac{8}{( s ^{2} + 6 s + 8 )}

\frac{d}{d x} (lo g_{10} ((arcsin (x))^{2}))

\int_{- 1}^{1} (x^{2} - 2 x + 3) d x

What is the solution for x^2y^{''}+xy^'-y= 1/x ?
The solution for x^2y^{''}+xy^'-y= 1/x is y=c_{1}x+(c_{2})/x-1/(4x)-(ln(x))/(2x)