# Solve the initial value problem

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

1. To solve this, we should use the substitution
$u=$
With this substitution,
$y=$
$y'=$
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 $z(0)=36$.