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Topics - Jiwen Bi

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Quiz-5 / TUT 0202 QUIZ5
« on: November 01, 2019, 02:02:53 PM »
$y''+9y=9sec^{2}3t,for\,0< t< \frac{\pi}{6}\\
we\,solve\,homogeneous\,solution\,first\\
y''+9y=0\\
so\,r^{2}+9=0\\
r=\pm 3i\\
homogeneous\,solution\,:y_{c}(t)=c_{1}cos(3t)+c_{2}sin(3t)\\
now\,solve\,the\,general\,solution\\
sub\,c_{1},c_{2}\,to\,u_{1},u_{2}\\
y(t)=u_{1}(t)cos(3t)+u_{2}(t)sin(3t)\\
u_{1}(t)=-\int \frac{y_{2}(t)g(t)}{W(y_{1},y_{2})(t)}dt+c_{1}\\
u_{2}(t)=-\int \frac{y_{1}(t)g(t)}{W(y_{1},y_{2})(t)}dt+c_{2}\\
w(y_{1},y_{2})(t)=3(cos^{2}(3t)+sin^{2}(3t)=3\\
FOR\,y_{1}(t)=cos3t,y_{2}(t)=sin3t,g(t)=9sec^{2}(3t)\\
u_{1}(t)=-\int \frac{sin(3t)9sec^{2}(3t)}{3}dt+c_{1}=-sec(3t)+c_{1}\\
u_{2}(t)=-\int \frac{cos(3t)9sec^{2}(3t)}{3}dt+c_{2}=ln(tan(3t)+sec(3t))
+c_{2}\\
y(t)=homogeneous eend ous+particular\\
y(t)=c_{1}cos(3t)+c_{2}sin(3t)+(ln(tan(3t)+sec(3t))+c_{2})sin(3t)-1
w(y_{1},y_{2})(t)=3(cos^{2}(3t)+sin^{2}(3t)=3\\
u_{1}(t)=-\int \frac{sin(3t)9sec^{2}(3t)}{3}dt+c_{1}=-sec(3t)+c_{1}\\
u_{2}(t)=-\int \frac{cos(3t)9sec^{2}(3t)}{3}dt+c_{2}=ln(tan(3t)+sec(3t))
+c_{2}\\
y(t)=homogeneous eend ous+particular\\
y(t)=c_{1}cos(3t)+c_{2}sin(3t)+(ln(tan(3t)+sec(3t))+c_{2})sin(3t)-1
$

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Quiz-4 / Tut 0202 QUIZ 4
« on: October 18, 2019, 07:51:06 PM »
$The\,Chaacteristic\,eqution\,is:\\
4r^{2}+12r+9=0\\
the\,roots\,of\,this\,eqution\,is\\
r_{1,2}=\frac{-12\pm \sqrt{144-16*9}}{8}\\
=\frac{-12\pm 0}{8}\\
=-\frac{3}{2}\\
Result:y(t)=c_{1}e^{-\frac{3t}{2}}+c_{2}te^{-\frac{3}{2}}$

3
Quiz-3 / TUT0202 Quiz 3
« on: October 11, 2019, 02:00:10 PM »
$2y''-3y'+y=0\\
The\,given\,differemtial\,eqution\,is\,\\
2y''-3y'+y=0\\
we\,assume\,that\,y=e^{rt}\,is\,a\,solution\,of\\2y''-3y'+y=0\\
Now \,y=e^{rt}\\
then\,y'=re^{rt}\\
y''=r^{2}e^{rt}\\
using\,these\,values\,in\,eqution\,2y''-3y'+y=0\\
2r^{2}e^{rt}-3re^{rt}+e^{rt}=0\\
e^{rt}(2r^2-3r+1)=0\,since\,e^{rt}\,\neq 0\\
2r^2-3r+1=0\,,2r(r-1)-1(r-1)=0\\
(r-1)(2r-1)=0\\
Then\,r=(1,\frac{1}{2})\\
then\,e^{t}and\,e^{\frac{1}{2}t}\,are\,two\,solution\,of\,eqution\,and\,hence\,the\,general\,solution\,is\,given\,by\\
y=c_{1}e^{t}+c_{2}e^{\frac{t}{2}}$

4
Quiz-2 / 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$

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