$2yy'+5=y^2 +5x$ with $y(0)=6$

  1. To solve this, we should use the substitution
    With this substitution,
  2. After the substitution from the previous part, we obtain the following linear differential equation in $x, u, u'$

  3. The solution to the original initial value problem is described by the following equation in $x,y$.
    y =sqrt{(36e^x)-5x}

I need a step-by-step $u$-substitution but have no idea how to work it out.
I found the answer on wolfram but I have no idea how to solve it.


Hint. Substitution $z=y^2$, and then you obtain the IVP

$z'-z=5x-5,$ with $z(0)=36$.


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