How to solve this second-order quasilinear partial differential equation?

Question

Now, I have a equation $f(x,y)$, satisfy the following PDE: $$ \frac{1}{2}f_{xx}-\frac{f_{x}}{2x}=\frac{f_{y}}{f}. $$ Now, putting constraint conditions aside, I wonder how could be the form of the solution $f(x,y)$. Could you help me?

Answer 1

Seems that a separable solution of the form $f(x,y) = X(x)Y(y)$ would work. Plugging this in, we have

$$\frac 12X''Y-\frac{1}{2x}X'Y = \frac{Y'}{Y}$$

Divide through by $Y(y)$:

$$\implies \frac 12 X'' - \frac{1}{2x}X' = \frac{Y'}{Y^2}$$

The LHS is a function of only $x$, whereas the RHS only $y$. Since they equate, they must each be equal to some constant $\lambda$:

$$\frac 12 X'' - \frac{1}{2x}X' = \frac{Y'}{Y^2} = \lambda$$

That is, we now have the two equations

$$\frac 12 X'' - \frac{1}{2x}X' = \lambda \qquad \qquad \text{and} \qquad \qquad \frac{Y'}{Y^2} = \lambda$$

which can be solved to give

$$X(x) = Ax^2+B+g(x,\lambda) \qquad \qquad \text{and} \qquad \qquad Y(y) = \frac{1}{C-\lambda y}$$

where $A,B,C$ are constants to be determined by boundary conditions, and $g(x,\lambda)$ is a particular solution that I haven't quite figured out yet.

The required solution is then

$$f(x,y) = \frac{Ax^2+B+g(x,\lambda)}{C-\lambda y}$$