What is the solution for x^2y^{''}+xy^'-4y=x^2+x ?

The solution for x^2y^{''}+xy^'-4y=x^2+x is y=c_{1}x^2+(c_{2})/(x^2)+(12x^2ln(x)-3x^2-16x)/(48)

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x^2y^{''}+xy^'-4y=x^2+x

x^{2} y^{^{''}} + x y^{^{'}} - 4 y = x^{2} + x

y = c_{1} x^{2} + \frac{c _{2}}{x ^{2}} + \frac{12 x ^{2} ln ( x ) - 3 x ^{2} - 16 x}{48}

Solution steps

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

Solve linear ODE:

y = c_{1} x^{2} + \frac{c _{2}}{x ^{2}} + \frac{12 x ^{2} ln ( x ) - 3 x ^{2} - 16 x}{48}

y = c_{1} x^{2} + \frac{c _{2}}{x ^{2}} + \frac{12 x ^{2} ln ( x ) - 3 x ^{2} - 16 x}{48}

What is the solution for x^2y^{''}+xy^'-4y=x^2+x ?
The solution for x^2y^{''}+xy^'-4y=x^2+x is y=c_{1}x^2+(c_{2})/(x^2)+(12x^2ln(x)-3x^2-16x)/(48)