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

The solution for x^2y^{''}+xy^'-y=(ln(x))/(x^2) is y=c_{1}x+(c_{2})/x+(3ln(x)+4)/(9x^2)

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x^2y^{''}+xy^'-y=(ln(x))/(x^2)

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

y = c_{1} x + \frac{c _{2}}{x} + \frac{3 ln ( x ) + 4}{9 x ^{2}}

Solution steps

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

Solve linear ODE:

y = c_{1} x + \frac{c _{2}}{x} + \frac{3 ln ( x ) + 4}{9 x ^{2}}

y = c_{1} x + \frac{c _{2}}{x} + \frac{3 ln ( x ) + 4}{9 x ^{2}}

\frac{x + 1}{x ^{2} + 1}

y^{^{'}} = 4 y + 1

\int_{2}^{4} \frac{6 - x - y}{8} d y

n = 1 \sum \infty \frac{n}{2 ^{n} + 5}

What is the solution for x^2y^{''}+xy^'-y=(ln(x))/(x^2) ?
The solution for x^2y^{''}+xy^'-y=(ln(x))/(x^2) is y=c_{1}x+(c_{2})/x+(3ln(x)+4)/(9x^2)