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### Messages - 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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##### Term Test 1 / Re: Problem 4 (noon)
« on: October 23, 2019, 09:28:40 PM »
$y''-8y'+25y=18e^{4x}+104cos(3x)\\ transfer\,it\,to\, homogenous \,eqution:\\ r^{2}-8r+25=0\\ Solve\,it\\ r^{2}-8r+16=-9\\ (r-4)^{2}=-9,r-4=9i\\ r_{1}=3i+4\\ r_{2}=-3i+4\\ y_{c}(x) solution:\\ y_{c}(x)=C_{1}e^{4x}cos{3x}+C_{2}e^{4x}sin{3x}\\ Now\,find\,first\,y_{p}(x)\\ y''-8y'+25y=18e^{4x}\\ y_{p}(x)=Ae^{4x}\\ y'=4Ae^{4x}\\ y'=16Ae^{4x}\,,then y''-8y'+25y=16Ae^{4x}-8(4Ae^{4x})+25(Ae^{4x})\\ A=2\\ Y_{P}(x)=2e^{4x}\\ find\,second\,y_{p}(x),\\ y''-8y'+25y=104cos(3x),\\ let \,y_{P}=Acos(3x)+Bsin(3x)\\ y'_{p}=-3Asin(3x)+3Bcos(3x)\\ y''_{p}=-9Acos(3x)-9Bsin(3x)\\ y''-8y'+25y=-9Acos(3x)-9Bsin(3x)-8(-3Asin(3x)+3Bcos(3x))+25(-3Asin(3x)+3Bcos(3x)\\ A=2,B=-3\,y_{P}=2cos(3x)-3sin(3x)\\ y=y_{c}+y_{p}=C_{1}e^{4x}cos{3x}+C_{2}e^{4x}sin{3x}+2e^{4x}+2cos(3x)-3sin(3x)\\ let\,y(0)=0,y'(0)=0,\\ C_{1}=-4,C_{2}=\frac{1}{3}\\ y(x)=-4e^{4x}cos{3x}+\frac{1}{3}e^{4x}sin{3x}+2e^{4x}+2cos(3x)-3sin(3x)$

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##### Term Test 1 / Re: Problem 3 (noon)
« on: October 23, 2019, 06:45:21 AM »
${y}''-{4y}'+3y=96sinh(x)\\\\ (a). Homogeneous part:\\ r^2-4r+3=96sinh(x)\\\\ r^2-4r+3=96(\frac{e^{x}-e^{-x}}{2})=48e^{x}-48e^{x}\\\\ let\,r^2-4r+3=0\\ =(r-1)(r-3),r_{1}=1,r_{2}=3\\ y_{c}=C_{1}e^{x}+C_{2}e^{3x}\\ Next\,we\,solve \, {y}''+{4y}'+3y=48e^{x}\\ Since \,we \,already \,have \,Ae^{x} in \,p_{c} \,solution\\ let y_{p}=Axe^{x} then\,{y}'=Axe^{x}+Ae^{x},\,{y}''=2Ae^{x}+Axe^{x}\\ {y}''+{4y}'+3y=2Ae^{x}+Axe^{x}+4(Axe^{x}+Ae^{x})+3(Axe^{x})\\ -2Ae^{x}=48e^{x}\\ -2A=48 A=-24,y_{p}=-24e^{x}\\ now\,let\,{y}''-{4y}'+3y=-48e^{-x}\\ let\, y_{p}=Be^{-x}\\ then\,{y}'=-Be^{-x},{y}''=Be^{-x}\\ {y}''-{4y}'+3y=Be^{-x}-4(-Be^{-x})+3(Be^{-x})\\ 8Be^{x}=-48e^{x}\\ B=-6 y_{P}=-6e^{x}\\ So\, the \,general \,solution \,is\,y=y_{c}+y_{P}=C_{1}e^{x}+C_{2}e^{3x}-24xe^{x}-6e^{-x}\\\\ (b)y=C_{1}e^{x}+C_{2}e^{3x}-24xe^{x}-6e^{-x}\\ {y}'=C_{1}e^{x}+3C_{2}e^{3x}-24e^{x}-24xe^{x}+6e^{-x}\\ let\,y(0)=0\\C_{1}+C_{2}-6=0,C_{1}+C_{2}=6\\ let\,y(0)'=0 C_{1}+3C_{2}=18\\ C_{1}= 0,C{2}=6\\ so,\,y=6e^{3x}-24xe^{x}-6e^{-x}$

OK. V.I.

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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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