What is the solution for xy^'=y+(xy)^{1/2} ?

The solution for xy^'=y+(xy)^{1/2} is y= 1/4 xln^2(x)+c_{1}xln(x)+c_{1}^2x

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xy^'=y+(xy)^{1/2}

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

y = \frac{1}{4} x ln^{2} (x) + c_{1} x ln (x) + c_{1}^{2} x

Solution steps

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

Solve Bernoulli ODE:

y = \frac{1}{4} x ln^{2} (x) + c_{1} x ln (x) + c_{1}^{2} x

y = \frac{1}{4} x ln^{2} (x) + c_{1} x ln (x) + c_{1}^{2} x

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What is the solution for xy^'=y+(xy)^{1/2} ?
The solution for xy^'=y+(xy)^{1/2} is y= 1/4 xln^2(x)+c_{1}xln(x)+c_{1}^2x