Hah, this was one of my assignment questions a couple of years ago. I'll just post it up in full.
Let us parametrise the characteristic curve in the plane of this partial differential equation by
,y(s)))
, and let
 = (x(s),y(s)))
. As the chain rule implies that
it follows that we can write
e^{x(s)} + y(s)u(s) , \qquad \frac{dx}{ds} = y(s) , \qquad \frac{dy}{ds} = x(s).<br />)
Writing
,y(s)) = \underline{z}(s))
, we see that the latter two ordinary differential equations are equivalent to the vector-valued ordinary differential equation
.<br />)
The matrix above has eigenvalues
with corresponding eigenvectors
)
, and so the solution to this ordinary differential equation is given by
 = c_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} e^{s} + c_2 \begin{pmatrix} 1 \\ -1 \end{pmatrix} e^{-s} ,<br />)
where
. Thus we have that
 = c_1 e^{s} + c_2 e^{-s})
and
 = c_1 e^{s} - c_2 e^{-s})
. Substituting these values into the earlier expression for
, we have that
e^{c_1 e^{s} + c_2 e^{-s}} + (c_1 e^{s} - c_2 e^{-s})u(s),<br />)
which, by multiplying through by the integrating factor
, can be rearranged to read
 = c_1 e^{s} + c_2 e^{-s},<br />)
and so we can integrate to find that
 = c_1 e^{s} - c_2 e^{-s} + u_0,<br />)
or equivalently, by substituting back in
,
, and then multiplying through by
,
 = e^{x}\left(y + u_0\right),<br />)
where
depends on the chosen characteristic curve. These characteristic curves are the solutions to
which, by separating variables, we can solve to find that
where
. Thus we have that
 = f(C))
for any arbitrary function
, and so
 = e^{x}\left(y + f\left(y^2 - x^2\right)\right).<br />)
With this solution of the partial differential equation, we therefore have that on the upper portion of the hyperbola
,
,
 = e^{x}\left(y + f(1)\right).<br />)
Thus if
is not of the form
 = e^{x}\left(y + \phi_0\right))
for some
, then we cannot have that
 = \phi(x,y))
on the upper portion of the hyperbola
,
. If, on the other hand,
is of this form, then there are infinitely many solutions of the partial differential equation satisfying this prescribed condition, as there are infinitely many functions
satisfying
 = \phi_0)
; for example, we could have
 = m\left(y^2 - x^2\right) - m + \phi_0)
for any
.
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