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

The solution for y^2=(xyy^'+1)ln(x) is y=sqrt(2ln(x)+c_{1)ln^2(x)},y=-sqrt(2ln(x)+c_{1)ln^2(x)}

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y^2=(xyy^'+1)ln(x)

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

y^{^{'}} = 1 - y

t^{2} y^{^{'}} + 6 t y - y^{3} = 0

y^{^{''}} - 2 y^{^{'}} + 4 y = 0

e^{x^{2} - 4 x}

8 e^{4 x}

What is the solution for y^2=(xyy^'+1)ln(x) ?
The solution for y^2=(xyy^'+1)ln(x) is y=sqrt(2ln(x)+c_{1)ln^2(x)},y=-sqrt(2ln(x)+c_{1)ln^2(x)}