What is the solution for y^{''}-2/x y^'-4/(x^2)y=3x^2 ?

The solution for y^{''}-2/x y^'-4/(x^2)y=3x^2 is y=c_{1}x^4+(c_{2})/x+3/5 x^4ln(x)-(3x^4)/(25)

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y^{''}-2/x y^'-4/(x^2)y=3x^2

y^{^{''}} - \frac{2}{x} y^{^{'}} - \frac{4}{x ^{2}} y = 3 x^{2}

y = c_{1} x^{4} + \frac{c _{2}}{x} + \frac{3}{5} x^{4} ln (x) - \frac{3 x ^{4}}{25}

Solution steps

y^{''} (x) - \frac{2}{x} y^{'} (x) - \frac{4}{x ^{2}} y = 3 x^{2}

Solve linear ODE:

y = c_{1} x^{4} + \frac{c _{2}}{x} + \frac{3}{5} x^{4} ln (x) - \frac{3 x ^{4}}{25}

y = c_{1} x^{4} + \frac{c _{2}}{x} + \frac{3}{5} x^{4} ln (x) - \frac{3 x ^{4}}{25}

x \to 1 + lim (x^{3} - 4)

x \to - 1 lim (\frac{2 x}{x + 1})

n = 1 \sum \infty \frac{3}{n ^{4}}

\frac{d}{d x} (\frac{1}{arctan ( x )})

h \to 0 lim (\frac{( 1 + h ) ^{2} - 1}{h})

What is the solution for y^{''}-2/x y^'-4/(x^2)y=3x^2 ?
The solution for y^{''}-2/x y^'-4/(x^2)y=3x^2 is y=c_{1}x^4+(c_{2})/x+3/5 x^4ln(x)-(3x^4)/(25)