"4Uxx+5Uxy+Uyy+Ux+Uy=0\\\\\nor, 4r+5s+t+p+q=0\\\\\nwhere R=4,S=5,T=1.\\\\\n\\text{therefore, characteristic equation is }R\\lambda^2+S\\lambda+T=0\\\\\n4\\lambda^2+5\\lambda+1=0\\\\\n\\lambda=-1\/4,-1\\\\\nThen,\\\\\n\\text{the characteristics curve of given pde is the solution of these equation}\\\\\n\\frac{dy}{dx}+\\lambda_1=0 --------(1)\\\\\nand\\\\\n\\frac{dy}{dx}-\\lambda_2=0---------(2)\n\\\\\n\\text{from (1),we get}\\\\\nc_1=y-(x\/4) (=m)\\\\\n\\text{from (2),we get}\\\\\nc_1=y-x (=n)\\\\\n\\text{Now, find p,q,r,s,t and substitute in the given pde.}\\\\\np=Ux=\\frac{-1}{4}Um-Un\\\\\nq=Uy=Um+Un\\\\\nr=Uxx=\\frac{1}{16}Umm+\\frac{1}{2}Umn+Unn\n\\\\\ns=\\frac{-1}{4}Umm-\\frac{5}{4}Umn-Unn\\\\\nt=Umm+2Umn+Unn\\\\\n\\text{substitute in the given pde, we get}\\\\\nUmm=\\frac{Um}{3}\\\\\n\\text{This is required canonical form.}"
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