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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int \frac{sec ( x ) tan ( x )}{sec ( x ) - 1} d x

(x + 2) (2 x^{2} - 3)

\int x e^{k x} d x

s + 1 - 2.82 j

\int 3 x^{2} - 18 d x

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