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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 »
$The\,Chaacteristic\,eqution\,is:\\ 4r^{2}+12r+9=0\\ the\,roots\,of\,this\,eqution\,is\\ r_{1,2}=\frac{-12\pm \sqrt{144-16*9}}{8}\\ =\frac{-12\pm 0}{8}\\ =-\frac{3}{2}\\ Result:y(t)=c_{1}e^{-\frac{3t}{2}}+c_{2}te^{-\frac{3}{2}}$

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

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##### 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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##### Quiz-1 / TUT0202 Quiz 1
« on: September 27, 2019, 03:35:10 PM »
${y}'=\frac{x^{2}+3y^{2}}{2xy}\\=\frac{x}{2y}+\frac{3y}{2x}=\frac{1}{2}(\frac{y}{x})^{-1}+\frac{3}{2}(\frac{y}{x})$

let  $v=\frac{y}{x}\\{v}'x+v={y}'\\ {v}'x+v=\frac{1}{2}v^{-1}+\frac{3}{2}v\\ {v}'x=\frac{1}{2}v^{-1}+\frac{1}{2}v=\frac{1+v^{2}}{2v}$
Equation is homogeneous
$\frac{2v}{1+v^{2}}dv=\frac{1}{x}\\ ln\left | v^{2}+1 \right |=ln\left | x \right |+c\\ v^{2}+1=cx\\ \frac{y^{2}}{x^{2}}+1-cx=0\\ y^{2}+x^{2}-cx^{3}=0$

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