Author Topic: TUT 0202 QUIZ2  (Read 315 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)$