What is the solution for x^2y^{''}+3xy^'+y= 1/x ?

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

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x^2y^{''}+3xy^'+y= 1/x

x^{2} y^{^{''}} + 3 x y^{^{'}} + y = \frac{1}{x}

y = \frac{c _{1}}{x} + \frac{c _{2} ln ( x )}{x} + \frac{ln ^{2} ( x )}{2 x}

Solution steps

x^{2} y^{''} (x) + 3 x y^{'} (x) + y = \frac{1}{x}

Solve linear ODE:

y = \frac{c _{1}}{x} + \frac{c _{2} ln ( x )}{x} + \frac{ln ^{2} ( x )}{2 x}

y = \frac{c _{1}}{x} + \frac{c _{2} ln ( x )}{x} + \frac{ln ^{2} ( x )}{2 x}

\frac{d}{d x} ((e^{x}) (cos (x)))

(6.1) (1.1)

(2 x + y + 1) (\frac{d y}{d x}) = 1

f (x) = \frac{32}{- 4}

\int_{3}^{4} 3 x cos (x^{2}) d x

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