solve the p.d.e-
r+3s+t-(s^2-rt)=1
by monge's method
where p,q,r,s,t have their usual meanings.
thnx
The Monge-Ampere equation
$\displaystyle R r+S s+T t+U\left(r t-s^2\right) = V$
was studied extensively in the 1800's most notably by Lie. He
showed that it was possible to find first integrals to this
equation. If one solves
$\displaystyle dp = r dx + s dy,\;\;\; dq = s dx + t dy$
for r and t and substitutes into our PDE and separate wrt s then
we obtain the Monge equations
$\displaystyle
R dy^2 - S dx dy + T dx^2 + U(dx dp + dy dq) =0,$
$\displaystyle
R dy dp + T dx dq + U dp dq - V dx dy = 0
$
Considering a linear combination of these two shows that they can
be factored if $\displaystyle \lambda$ are two real distinct solutions of
$\displaystyle
U^2 \lambda^2 + S U \lambda + TR + UV = 0
$ and further, the factors are
$\displaystyle
\left(U dp + T dx + \lambda_1 U dy\right)\left(U dq + \lambda_1 U dx + R dy \right) = 0
$
$\displaystyle
\left(U dp + T dx + \lambda_2 U dy\right)\left(U dq + \lambda_2 U dx + R dy \right) = 0
$
In your case where $\displaystyle R = 1, S = 3, T = 1, U = 1 \; \text{and}\; V = 1$ then we have $\displaystyle \lambda^2 + 3 \lambda + 2 = 0$ which has factors $\displaystyle \lambda = -1, -2$. Thus, we have
$\displaystyle \left(dp + dx - dy\right)\left( dq - dx + dy \right) = 0$
$\displaystyle \left( dp + dx - 2 dy\right)\left( dq - 2 dx + dy \right) = 0$
which gives rise to
$\displaystyle dp + dx - dy = 0, \;\; dq - 2 dx + dy = 0$
or
$\displaystyle
dp + dx - 2 dy = 0 ,\;\;\; dq - dx + dy =0
$
leading to the first integrals
$\displaystyle
p + x - y = f(q - 2x + y)
$ and
$\displaystyle p + x - 2y = g(q -x + y)$
Sophus Lie (1877) showed that via a contact transformation that it
possible to transform PDE like yours to the wave equation. In
your case the transformation is
$\displaystyle
x = U_X + U_Y,\;\;\; y = X - Y,\;\;\;u=(2X-Y)(U_X+U_Y) -
\frac{1}{2}(U_X+U_Y)^2-\frac{1}{2}(X-Y)^2-U
$ $\displaystyle \;\;\;(1)$
giving $\displaystyle U_{XY} = 0.$ With the general solution as $\displaystyle U = F(X) + G(Y)$, gives the general solution of your PDE via (1).