Paper 2, Section I, 5C5 \mathrm{C}5CMethodsPart IB, 2018Show thata(x,y)(dyds)2−2b(x,y)dxdsdyds+c(x,y)(dxds)2=0a(x, y)\left(\frac{d y}{d s}\right)^{2}-2 b(x, y) \frac{d x}{d s} \frac{d y}{d s}+c(x, y)\left(\frac{d x}{d s}\right)^{2}=0a(x,y)(dsdy)2−2b(x,y)dsdxdsdy+c(x,y)(dsdx)2=0along a characteristic curve (x(s),y(s))(x(s), y(s))(x(s),y(s)) of the 2nd 2^{\text {nd }}2nd -order pdea(x,y)uxx+2b(x,y)uxy+c(x,y)uyy=f(x,y)a(x, y) u_{x x}+2 b(x, y) u_{x y}+c(x, y) u_{y y}=f(x, y)a(x,y)uxx+2b(x,y)uxy+c(x,y)uyy=f(x,y)