Author Topic: TUT0202 Quiz 2  (Read 484 times)

Jiwen Bi

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TUT0202 Quiz 2
« on: October 04, 2019, 02:00:15 PM »
Jiwen Bi
$The\,required\,integrating\,factor\,is\, \mu =e^{3x}
solve\,the\, differential\,eqution(3x^{2}+2xy+y^{3})e^{3x}dx+(x^{2}+y^{2})dy=0 \\
there\,exist\,a\,function\,\psi(x,y)such\,that\, \psi_{x}(x,y)=M(x,y)and\,\psi_{y}(x,y)=N(x,y)\\
this\, implies \,\psi_{x}(x,y)=(3x^{2}+2xy+y^{3})e^{3x}an\,\psi_{y}(x,y)=(x^{2}+y^{2})e^{3x}\\
intefrate\,the\,first\,of\,above\,eqution\\
\psi_{x}(x,y)=(3x^{2}+2xy+y^{3})e^{3x}\\
\psi(x,y)=\int(3x^{2}+2xy+y^{3})e^{3x}dx\\
=3y\int x^{2}e^{3x}dx+2y\int xe^{3x}dx+e^{3x}dx\\
=3y(\frac{x^{2}e^{3x}}{3}-\frac{2xe^{3x}}{9}+\frac{2e^{3x}}{27})+\frac{2y}{3}(xe^{3x}-\frac{e^{3x}}{3})+\frac{e^{3x}}{3}y^{3}+h(y)
=x^{2}ye^{3x}+\frac{y^{3}}{3}e^{3x}+h(y)\\
differentiate \,above eqution\, with \,respect \,to\, x \,and \,equate\,to\,N\\
\psi(x,y)=x^{2}ye^{3x}+\frac{y^{3}}{3}e^{3x}+h(y)\\
\psi_{y}(x,y)=x^{2}e^{3x}+y^{2}e^{3x}+{h}'(y)\\
set \,\psi_{y}=N\,as\,follows:\\
x^{2}e^{3x}+y^{2}e^{ex}+{h}'(y)=x^{2}e^{3x}+y^{2}e^{3x}\\
{h}'(y)=0\\
this\,implies\,h(y)=0\\
substitude\,h(y)=0\,in\,equation\,\psi(x,y)=x^{2}ye^{3x}+\frac{y^{3}}{3}e^{3x}+h(y).\\

\psi(x,y)=x^{2}ye^{3x}+\frac{y^{3}}{3}e^{3x}+0\\
=x^{2}ye^{3x}+\frac{y^{3}}{3}e^{3x}\\
then\,the\,solution\,of\,differential\,eqution\,is\\
x^{2}ye^{3x}+\frac{y^{3}}{3}e^{3x}=k\\
3x^{2}ye^{3x}+y^{3}e^{3x}=3k\\
(3x^{2}y+y^{3})e^{3x}=C  {3k = C}\\
Hence,\,the\,required\,solution\,of\,the\,differential\,eqution\,is(3x^{2}y+y^{3})e^{3x}=C$
« Last Edit: October 11, 2019, 01:34:20 PM by Jiwen Bi »