### Author Topic: TUT 0202 QUIZ2  (Read 729 times)

#### Yijin Qiang

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##### TUT 0202 QUIZ2
« on: October 04, 2019, 02:00:01 PM »
$Now, \, find \, M_{y}(x,y)\\ =\frac{\delta }{\delta y} (\frac{x}{(x^{2}+y^{2})^{\frac{3}{2}}}\\ =x\frac{\delta }{\delta y}(x^{2}+y^{2})^{\frac{3}{2}}\\ =x(-\frac{3}{2})(x^{2}+y^{2})^{-\frac{5}{2}}(2y)\\ =-(\frac{3yx}{(x^{2}+y^{2})^{\frac{5}{2}}}\\ and \,also, \, N_{x}(x,y)=\frac{\delta }{\delta x}N(x,y)\\ =\frac{\delta }{\delta x} (\frac{y}{(x^{2}+y^{2})^{\frac{3}{2}}})\\ =y\frac{\delta }{\delta x}(x^{2}+y^{2})^{-{\frac{3}{2}}}\\ =y\frac{\delta }{\delta x}{-\frac{3}{2}}((x^{2}+y^{2})^{-{\frac{3}{2}}})\\ =-\frac{3yx}{{x^{2}+y^{2}}^{\frac{2}{5}}} here, \,observe \,that \,M_{y}(x,y)=N_{x}(x,y)=-\frac{3yx}{({x^{2}+y^{2}})^{\frac{5}{2}}}\\ so \,,the \,given \,differential \,equation \,is \,exact.\\ since \,the \,given \,eqution \,is \,exact \,, then \,there \,exists \,solution,\psi (x,y)such\,that\\ \psi_{x}(x,y)=M(x,y)\\ \psi_{x}(x,y)=\frac{x}{(x^{2}+y^{2})^{\frac{3}{2}}}\\ integrate\,with\,respect\,to\,x, \,then \int \psi_{x}(x,y)=\int \frac{x}{(x^{2}+y^{2})^{\frac{3}{2}}}dx\\ =\frac{1}{2}\int(x^{2}+y^{2})^{-{\frac{3}{2}}}(2x)dx\\ \frac{1}{2}\frac{({x^{2}+y^{2}})^{-{\frac{1}{2}}}}{-{\frac{1}{2}}}\\ \psi(x,y)=-\frac{1}{{(x^{2}+y^{2})}^{\frac{1}{2}}}+h(y) \\ (2)\\ next\,find \psi_{y} from (2)\\ differentiate\,it\,with\,respect\,to\,y, \,then \\ \psi_{y}(x,y)=\frac{\delta }{\delta y}(\psi(x,y))\\ =\frac{\delta }{\delta y}(-{\frac{1}{(x^{2}+y^{2})^{\frac{1}{2}}}}+h(y))\\ =-\frac{\delta }{\delta y}{\frac{1}{(x^{2}+y^{2})^{\frac{1}{2}}}}+{\frac{\delta}{\delta y}}h(y)\\ =\frac{1}{{(x^{2}+y^{2})}^{\frac{3}{2}}}+{h}'(y)\\ set\,the\,result\,equal\,to\,N(x,y)\\ \frac{y}{{(x^{2}+y^{2})}^{\frac{3}{2}}}+{h}'(y)=\psi_{x}(x,y)\\ \frac{y}{{(x^{2}+y^{2})}^{\frac{3}{2}}}+{h}'(y)=N(x,y)\\ \frac{y}{{(x^{2}+y^{2})}^{\frac{3}{2}}}+{h}'(y)=\frac{y}{{(x^{2}+y^{2})}^{\frac{3}{2}}}\\ {h}'(y)=0\\ integrate\,with\,respect\,to\,y, then\\ {h}'(y)=0\\ h(y)=0\\ now\,,substitute\,the\,value\,of\,h(y) \,in\,equation(2)\\ \psi(x,y)=-\frac{1}{{(x^{2}+y^{2})}^{\frac{1}{2}}}+h(y)\\ \psi(x,y)=-\frac{1}{{(x^{2}+y^{2})}^{\frac{1}{2}}}+0\\ \psi(x,y)=-\frac{1}{{(x^{2}+y^{2})}^{\frac{1}{2}}}\\ then\,the\,solutions\,of\,equations(1) \,are\,given\,implicitly\,by \\ -\frac{1}{{(x^{2}+y^{2})}^{\frac{1}{2}}}=k\\ {(x^{2}+y^{2})}^{\frac{1}{2}}=\frac{-{1}}{k}\\ x^{2}+y^{2}=\frac{1}{k^{2}}\\ Hence,\,the\,required\,solution\,is\,x^{2}+y^{2}=c,\,here\\ c =\frac{1}{k^{2}}(arbitrary\,constant)$