Another attempt at solving a PDE with the method of characteristics I want to use the method of characteristics to obtain the solution to this PDE,
$$\frac{\partial F}{\partial t}=\left(z-t\right)\left(\beta z-\gamma\right)\frac{\partial F}{\partial z}$$
which I've seen is of the form
$$F\left(t,z\right)=F\left(\left(\frac{\beta\left(z-1\right)}{\beta z-\gamma}\right)e^{\left(\beta-\gamma\right)t}\right).$$
Attempt:
I was trying by identifying the PDE with one of the form
$$a\left(t,z\right)F_{z}+b\left(t,z\right)F_{t}=g\left(t,z,F\right),$$
deriving the solution $h\left(t,z\right)=c_{1}$ for
$$\frac{dz}{dt}=\frac{a\left(t,z\right)}{b\left(t,z\right)}$$
then the solution $j\left(t,z,F\right)=c_{2}$ for
$$\frac{dF}{dt}=\frac{g\left(t,z,F\right)}{b\left(t,z\right)}$$
and writting $j\left(t,z,F\right)=K\left(h\left(t,z\right)\right)$.
However, I got stuck when I stomped upon the following ODE to solve,
$$z'\left(t\right)=-\beta z^{2}\left(t\right)+\beta tz\left(t\right)+\gamma z\left(t\right)-\gamma t.$$
How can I solve the problem?
 A: Riccati's equation $$z'\left(t\right)=-\beta z^{2}\left(t\right)+\beta tz\left(t\right)+\gamma z\left(t\right)-\gamma t$$
$$z=-\beta z(z-t)+\gamma( z- t)$$
$$z=(\gamma-\beta z)( z- t)$$
Substitute $u=(\gamma-\beta z) \implies z=\frac {(\gamma -u)} {\beta}$
$$-\frac {u'}{\beta}=u(\frac {\gamma}{\beta}-\frac u {\beta}-t)$$
$$u'=u( {u-\gamma} + t {\beta})$$
Which is simply Bernouilli's equation ...
$$\boxed{u'=u(t {\beta}-\gamma)+ u^2}$$
$$-(\frac 1 u)'= \frac {(t {\beta}-\gamma)} u+ 1 $$
$$\boxed{w'= w (\gamma-t\beta)-1 \text{, where } w=\frac 1 u=\frac 1 {\gamma-\beta z} }$$
And thats a first ODE easy to solve...
A: $$z'\left(t\right)=-\beta z^{2}\left(t\right)+(\beta t+\gamma)z\left(t\right)-\gamma t.$$
Let $\quad z(t)=\frac{y'(t)}{\beta y(t)} \quad\to\quad z'=\frac{y''}{\beta y}-\frac{y'^2}{\beta y^2} =-\beta \left(\frac{y'}{\beta y}  \right)^{2}+(\beta t+\gamma)\frac{y'}{\beta y} -\gamma t$
$$y''-(\beta t+\gamma)y' +\gamma\beta ty=0$$
This is a second order linear ODE easy to solve for $y(t)$ . Then, compute $z(t)=\frac{y'(t)}{\beta y(t)} $.
A: Just to clarify the function substitution in Jaquelins answer: the substitution is related to the Logarithmic derivative which is very famouse : $$\frac{\partial \{\log(y(t))\}}{\partial t} = \frac{y'(t)}{y(t)}$$ Where $y'(t)$ is the inner derivative and $\frac{\partial\log(y)}{\partial y} = \frac{1}y$ the outer derivative according to the Chain rule.
