Background: I've solved this linear PDE: $$\frac{\partial H_a}{\partial p_r}-g_0({p_r})\frac{\partial H_a}{\partial z}=0$$ Like this (and this is true) that I used method of characteristics: $$\frac{dp_r}{ds}=1 \And (p_r(0)=0)\implies p_r=s$$ $$\frac{dz}{ds}=g_0({p_r}) \And(z(0)=\aleph)\implies z=-\int_{0}^{p_r} g_0(s) ds+\aleph$$ $$\frac{dH_a}{ds}=0 \And (H_a(0)=\phi(\aleph))\implies H_a(z,p_r)=\phi(z+\int_{0}^{p_r} g_0(s) ds)$$
Question: Now, I should solve this quasilinear PDE:
$$\frac{\partial H_a}{\partial p_r}-2g_0({p_r})\frac{\partial H_a}{\partial z}=g_0({p_r})k_0z$$
Although I like to use method of characteristics again, I don't know hot to do. So I do this:
Forming this: $$\frac{dp_r}{1}=\frac{dz}{-2g_0({p_r})}=\frac{dH_a}{g_0({p_r})k_0z}$$ Taking first two fractions integral leads: $$dz=-2g_0({p_r})dp_r \implies z=-2\int_{0}^{p_r} g_0(s) ds+c_1$$ And for last two fractions: $$\frac{k_0}{-2}zdz=dH_a \implies H_a=\frac{k_0}{-2}z^2+c_2$$
And I'm sure I miss some points (for example $c_1$ and $c_2$ are not necessarily constants). Could anyone help me?