I'm solving for the equilibrium strategy in a first-price auction with common value. During the process of maximization of the utility function of a generic bidder I've encountered the following differential equation, but I'm not able even to guess a possible solution because I've never studied ODE.
Start from here:
$$ \pi= \int_{0}^{\omega}v + y- b(\omega)-b(\omega)+ydy$$
then
$$\frac{d \pi}{d \omega} = v + 2 \omega -2[b(\omega)+ \omega b'(\omega)]$$
My problem is to solve the following:
$$3v - 2b(v) - 2 vb'(v) = 0 $$
where $ b(\cdot)$ is a strictly increasing function and $b(0)$ need not to be equal to $0$. Does somebody can help me?
Hint
$$3v - 2b(v) - 2 vb'(v) = 0$$ $$3v - 2(b + vb') = 0$$ $$3v - 2(bv)' = 0$$ $$ (bv)'= \frac 3 2 v $$ Simply integrate : $$ bv=\frac 3 2 \int vdv $$ For $v \neq 0 $ $$\boxed { b(v)=\frac 3 4 v+ \frac Kv}$$ Note that for $(v=0)$ we have $3v - 2b(v) - 2 vb'(v) =-2b(0) =0$ And $b(0)=0$