What is the solution for y^{''}-3/x y^'-(12)/(x^2)y=4x^2 ?

The solution for y^{''}-3/x y^'-(12)/(x^2)y=4x^2 is y=c_{1}x^6+(c_{2}}{x^2}-\frac{x^4)/3

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y^{''}-3/x y^'-(12)/(x^2)y=4x^2

y^{^{''}} - \frac{3}{x} y^{^{'}} - \frac{12}{x ^{2}} y = 4 x^{2}

y = c_{1} x^{6} + \frac{c _{2}}{x ^{2}} - \frac{x ^{4}}{3}

Solution steps

y^{''} (x) - \frac{3}{x} y^{'} (x) - \frac{12}{x ^{2}} y = 4 x^{2}

Solve linear ODE:

y = c_{1} x^{6} + \frac{c _{2}}{x ^{2}} - \frac{x ^{4}}{3}

y = c_{1} x^{6} + \frac{c _{2}}{x ^{2}} - \frac{x ^{4}}{3}

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

\int 6 x^{2} e^{6 x} d x

y = x^{3} + x^{2} - 6 x, y = 6 x

e^{(3 t)} (4 cos (3 t) - \frac{10}{3} sin (3 t))

x \to - 1 + lim (\frac{x}{∣ x + 1 ∣})

What is the solution for y^{''}-3/x y^'-(12)/(x^2)y=4x^2 ?
The solution for y^{''}-3/x y^'-(12)/(x^2)y=4x^2 is y=c_{1}x^6+(c_{2}}{x^2}-\frac{x^4)/3