What is the solution for x^2y^{''}-3xy^'+4y=x ?

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

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x^2y^{''}-3xy^'+4y=x

x^{2} y^{^{''}} - 3 x y^{^{'}} + 4 y = x

y = c_{1} x^{2} + c_{2} ln (x) x^{2} + x

Solution steps

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

Solve linear ODE:

y = c_{1} x^{2} + c_{2} ln (x) x^{2} + x

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