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

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

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

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

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

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

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

x \to 0 lim (\frac{cos ( \frac{π}{2} - x )}{x})

n = 2 \sum \infty 5 \cdot (- \frac{3}{4})^{n}

y^{^{''}} + 5 y^{^{'}} + 4 y = t^{2} e^{7 t}

\int_{0}^{l n (5)} e^{x} - e^{- 2 x} d x

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