What is the solution for x^2y^{''}-2y=3x^2-1 ?

The solution for x^2y^{''}-2y=3x^2-1 is y=c_{1}x^2+(c_{2})/x+x^2ln(x)+1/6-(x^2-1)/3

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x^2y^{''}-2y=3x^2-1

x^{2} y^{^{''}} - 2 y = 3 x^{2} - 1

y = c_{1} x^{2} + \frac{c _{2}}{x} + x^{2} ln (x) + \frac{1}{6} - \frac{x ^{2} - 1}{3}

Solution steps

x^{2} y^{''} (x) - 2 y = 3 x^{2} - 1

Solve linear ODE:

y = c_{1} x^{2} + \frac{c _{2}}{x} + x^{2} ln (x) + \frac{1}{6} - \frac{x ^{2} - 1}{3}

y = c_{1} x^{2} + \frac{c _{2}}{x} + x^{2} ln (x) + \frac{1}{6} - \frac{x ^{2} - 1}{3}

y = ln (2 - 9 x)

\frac{d}{d x} (cos (3 x^{5} + 1))

x \to \infty lim (x^{2} + y)

x y^{^{'}} + 3 y = \frac{e ^{x}}{x}

\int \frac{1}{( x ^{2} - 36 ) ^{2}} d x

What is the solution for x^2y^{''}-2y=3x^2-1 ?
The solution for x^2y^{''}-2y=3x^2-1 is y=c_{1}x^2+(c_{2})/x+x^2ln(x)+1/6-(x^2-1)/3