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

The solution for x^2y^'+2xy=ln(x),y(1)=7 is y=(ln(x))/x-1/x+8/(x^2)

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x^2y^'+2xy=ln(x),y(1)=7

x^{2} y^{^{'}} + 2 x y = ln (x), y (1) = 7

y = \frac{ln ( x )}{x} - \frac{1}{x} + \frac{8}{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{8}{x ^{2}}

y = \frac{ln ( x )}{x} - \frac{1}{x} + \frac{8}{x ^{2}}

y = 4 + x

\frac{d ^{2} y}{d x ^{2}} + 8 \frac{d y}{d x} + 16 y = 0

cos (π x), 0

\frac{16}{x}

\frac{\partial}{\partial y} (3 x^{2})

What is the solution for x^2y^'+2xy=ln(x),y(1)=7 ?
The solution for x^2y^'+2xy=ln(x),y(1)=7 is y=(ln(x))/x-1/x+8/(x^2)