What is the solution for x^2y^'+2xy=ln(x),y(1)=8 ?

The solution for x^2y^'+2xy=ln(x),y(1)=8 is y=(ln(x))/x-1/x+9/(x^2)

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x^2y^'+2xy=ln(x),y(1)=8

x^{2} y^{^{'}} + 2 x y = ln (x), y (1) = 8

y = \frac{ln ( x )}{x} - \frac{1}{x} + \frac{9}{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{9}{x ^{2}}

y = \frac{ln ( x )}{x} - \frac{1}{x} + \frac{9}{x ^{2}}

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What is the solution for x^2y^'+2xy=ln(x),y(1)=8 ?
The solution for x^2y^'+2xy=ln(x),y(1)=8 is y=(ln(x))/x-1/x+9/(x^2)