What is the solution for x^2y^'+2xy=ln(x),y(1)=6 ?

The solution for x^2y^'+2xy=ln(x),y(1)=6 is y=(ln(x))/x-1/x+7/(x^2)

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x^2y^'+2xy=ln(x),y(1)=6

x^{2} y^{^{'}} + 2 x y = ln (x), y (1) = 6

y = \frac{ln ( x )}{x} - \frac{1}{x} + \frac{7}{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{7}{x ^{2}}

y = \frac{ln ( x )}{x} - \frac{1}{x} + \frac{7}{x ^{2}}

x \to - 2 - lim (\frac{x}{x ^{2} - 4})

\frac{y}{2}

\frac{d}{d x} (- \frac{48 x}{( x ^{2} + 12 ) ^{2}})

\int e^{i x} d x

\frac{1}{sin ( x )}

What is the solution for x^2y^'+2xy=ln(x),y(1)=6 ?
The solution for x^2y^'+2xy=ln(x),y(1)=6 is y=(ln(x))/x-1/x+7/(x^2)