What is the solution for x^2y^'+2xy=ln(x),y(1)=3 ?

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

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x^2y^'+2xy=ln(x),y(1)=3

x^{2} y^{^{'}} + 2 x y = ln (x), y (1) = 3

y = \frac{ln ( x )}{x} - \frac{1}{x} + \frac{4}{x ^{2}}

Solution steps

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

Solve linear ODE:

y = \frac{ln ( x )}{x} - \frac{1}{x} + \frac{4}{x ^{2}}

y = \frac{ln ( x )}{x} - \frac{1}{x} + \frac{4}{x ^{2}}

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