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https://www.mathhomeworkanswers.org/289987/solve-the-following-by-monges-method-xy-t-r-x-2-y-2-s-2-py-qx
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For any z(x,y):
dz=(∂z/∂x)dx+(∂z/∂y)dy=pdx+qdy, from the definitions of p and q.
In the form Rr+Ss+Tt=V the given PDE becomes:
ry-sy+sx-tx+q-p=0⇒R=y, S=-y+x, T=-x, V=p-q.
r=(dp-sdy)/dx and t=(dq-sdx)/dy (see other question).
Therefore we can replace r and t:
y(dp-sdy)/dx+s(x-y)-x(dq-sdx)/dy=p-q, multiplying through by dxdy:
ydy(dp-sdy)+s(x-y)dxdy-xdx(dq-sdx)=(p-q)dxdy.
Monge's subsidiary equations are (see other question):
Rdydp+Tdxdq-Vdxdy=0 and Rdy2-Sdxdy+Tdx2=0, that is:
ydydp-xdxdq-(p-q)dxdy=0 and y(dy)2-(x-y)dxdy-x(dx)2=0.
The second of these equations can be written:
(dy)2-(x/y-1)dxdy-(x/y)(dx)2=0. We would like to factorise it:
(dy-m1dx)(dy-m2dx)=(dy)2-(m1+m2)dxdy+m1m2(dx)2≡(dy)2-(x/y-1)dxdy-(x/y)(dx)2=0.
So m1+m2=x/y-1 and m1m2=-x/y, and m1=x/y and m2=-1.
This gives us two possible equations:
dy-xdx/y=0⇒ydy-xdx=0; and dy+dx=0, both of which are integrable:
(1) y2/2-x2/2=k, y2-x2=a, a constant, y=√(x2+a), dy=xdx/√(x2+a);
(2) y+x=b, another constant, y=b-x; dy=-dx.
We can now make some substitutions in ydydp-xdxdq-(p-q)dxdy=0 to eliminate y:
(1)⇒(1.1) xdxdp-xdxdq-(p-q)(dx)2x/√(x2+a)=0, which reduces to:
dp-dq-(p-q)dx/√(x2+a)=0,
(dp-dq)/(p-q)=dx/(√(x2+a)) which is integrable:
Let a=c2, then ln|p-q|=(1/c)tan-1(x/c), or p-q=e(1/c)tan⁻¹(x/c). We can write this as p-q=f(x).
(2)⇒(2.1) -(b-x)dxdp-xdxdq+(p-q)(dx)2=0,
-(b-x)dp-xdq+(p-q)dx=0,
-bdp+xdp-xdq+pdx-qdx=0,
-bdq+d(px)-d(qx)=0 which is integrable:
-bq+px-qx=d, a constant.
d=g(b), where g is an arbitrary function.
-bq+px-qx=g(x+y),
We have -bq+px-qx=g(x+y) and p-q=f(x), i.e., p=q+f(x).
-bq+qx+xf(x)-qx=g(x+y), q=(xf(x)-g(x+y))/b, p=(bf(x)+xf(x)-g(x+y))/b.
dz=pdx+qdy=[bf(x)+xf(x)-g(x+y))/b]dx+[(xf(x)-g(x+y))/b]dy, the integral of which can be written:
z=F(x,y)+G(x,y)+H, where H is a constant and F=∫[bf(x)+xf(x)-g(x+y))/b]dx, G=∫[(xf(x)-g(x+y))/b]dy.
More specifically:
(p-q)x=bq, p-q=bq/x, p=q+qb/x=q(1+b/x).
So bq/x=e(1/c)tan⁻¹(x/c), q=(x/b)e(1/c)tan⁻¹(x/c),
so p=(1+b/x)(x/b)e(1/c)tan⁻¹(x/c),
p=((x+b)/b)e(1/c)tan⁻¹(x/c).
So p and q are each defined as functions of x.
dz=pdx+qdy=((x+b)/b)e(1/c)tan⁻¹(x/c)dx+(x/b)e(1/c)tan⁻¹(x/c)dy,
dz=e(1/c)tan⁻¹(x/c)((x+b)dx+xdy)/b, from which z can be derived.