What is the solution for 2x^2y^{''}+5xy^'+y=x^2-x ?

The solution for 2x^2y^{''}+5xy^'+y=x^2-x is y=(c_{1}}{sqrt(x)}+(c_{2})/x+\frac{x^2)/(15)-x/6

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2x^2y^{''}+5xy^'+y=x^2-x

2 x^{2} y^{^{''}} + 5 x y^{^{'}} + y = x^{2} - x

y = \frac{c _{1}}{x} + \frac{c _{2}}{x} + \frac{x ^{2}}{15} - \frac{x}{6}

Solution steps

2 x^{2} y^{''} (x) + 5 x y^{'} (x) + y = x^{2} - x

Solve linear ODE:

y = \frac{c _{1}}{x} + \frac{c _{2}}{x} + \frac{x ^{2}}{15} - \frac{x}{6}

y = \frac{c _{1}}{x} + \frac{c _{2}}{x} + \frac{x ^{2}}{15} - \frac{x}{6}

\frac{s}{( s ^{2} + 2 s + 2 )}

y^{^{'}} - \frac{1}{t} y = 0

\frac{d y}{d x} = x (7 - y)

n = 0 \sum \infty sin (πn)

\frac{- 4 x}{( x ^{2} - 1 ) ^{2}}

What is the solution for 2x^2y^{''}+5xy^'+y=x^2-x ?
The solution for 2x^2y^{''}+5xy^'+y=x^2-x is y=(c_{1}}{sqrt(x)}+(c_{2})/x+\frac{x^2)/(15)-x/6