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### Messages - Yijin Qiang

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1
##### Quiz-5 / TUT 0202 QUIZ5
« on: November 01, 2019, 02:02:31 PM »
$y''+9y=9sec^{2}3t,0< t< \frac{\pi}{6}\\ solve\,homogeneous\,solution\\ y''+9y=0\\ r^{2}+9=0\\ r=\pm 3i\\ y_{c}(t)=c_{1}cos(3t)+c_{2}sin(3t)\\ solve\,the\,general\,solution\\ replace\,c_{1},c_{2}\,with\,u_{1},u_{2}\\ y(t)=u_{1}(t)cos(3t)+u_{2}(t)sin(3t)\\ u_{1}(t)=-\int \frac{y_{2}(t)g(t)}{w(y_{1},y_{2})(t)}dt+c_{1}\\ u_{2}(t)=-\int \frac{y_{1}(t)g(t)}{w(y_{1},y_{2})(t)}dt+c_{2}\\ y_{1}(t)=cos3t,y_{2}(t)=sin3t,g(t)=9sec^{2}(3t)\\ w(y_{1},y_{2})(t)=3(cos^{2}(3t)+sin^{2}(3t)=3\\ u_{1}(t)=-\int \frac{sin(3t)9sec^{2}(3t)}{3}dt+c_{1}=-sec(3t)+c_{1}\\ u_{2}(t)=-\int \frac{cos(3t)9sec^{2}(3t)}{3}dt+c_{2}=ln(tan(3t)+sec(3t)) +c_{2}\\ y(t)=homogeneous eend ous+particular\\ y(t)=c_{1}cos(3t)+c_{2}sin(3t)+(ln(tan(3t)+sec(3t))+c_{2})sin(3t)-1 w(y_{1},y_{2})(t)=3(cos^{2}(3t)+sin^{2}(3t)=3\\ u_{1}(t)=-\int \frac{sin(3t)9sec^{2}(3t)}{3}dt+c_{1}=-sec(3t)+c_{1}\\ u_{2}(t)=-\int \frac{cos(3t)9sec^{2}(3t)}{3}dt+c_{2}=ln(tan(3t)+sec(3t)) +c_{2}\\ y(t)=homogeneous eend ous+particular\\ y(t)=c_{1}cos(3t)+c_{2}sin(3t)+(ln(tan(3t)+sec(3t))+c_{2})sin(3t)-1$

2
##### Quiz-5 / TUT 0202 QUIZ5
« on: November 01, 2019, 01:59:59 PM »
$y''+9y=9sec^{2}3t,0< t< \frac{\pi}{6}\\ solve\,homogeneous\,solution\\ y''+9y=0\\ r^{2}+9=0\\ r=\pm 3i\\ y_{c}(t)=c_{1}cos(3t)+c_{2}sin(3t)\\ solve\,the\,general\,solution\\ replace\,c_{1},c_{2}\,with\,u_{1},u_{2}\\ y(t)=u_{1}(t)cos(3t)+u_{2}(t)sin(3t)\\ u_{1}(t)=-\int \frac{y_{2}(t)g(t)}{w(y_{1},y_{2})(t)}dt+c_{1}\\ u_{2}(t)=-\int \frac{y_{1}(t)g(t)}{w(y_{1},y_{2})(t)}dt+c_{2}\\ y_{1}(t)=cos3t,y_{2}(t)=sin3t,g(t)=9sec^{2}(3t)\\ w(y_{1},y_{2})(t)=3(cos^{2}(3t)+sin^{2}(3t)=3\\ u_{1}(t)=-\int \frac{sin(3t)9sec^{2}(3t)}{3}dt+c_{1}=-sec(3t)+c_{1}\\ u_{2}(t)=-\int \frac{cos(3t)9sec^{2}(3t)}{3}dt+c_{2}=ln(tan(3t)+sec(3t)) +c_{2}\\ y(t)=homogeneous eend ous+particular\\ y(t)=c_{1}cos(3t)+c_{2}sin(3t)+(ln(tan(3t)+sec(3t))+c_{2})sin(3t)-1 w(y_{1},y_{2})(t)=3(cos^{2}(3t)+sin^{2}(3t)=3\\ u_{1}(t)=-\int \frac{sin(3t)9sec^{2}(3t)}{3}dt+c_{1}=-sec(3t)+c_{1}\\ u_{2}(t)=-\int \frac{cos(3t)9sec^{2}(3t)}{3}dt+c_{2}=ln(tan(3t)+sec(3t)) +c_{2}\\ y(t)=homogeneous eend ous+particular\\ y(t)=c_{1}cos(3t)+c_{2}sin(3t)+(ln(tan(3t)+sec(3t))+c_{2})sin(3t)-1$

3
##### Term Test 1 / Re: Problem 3 (main)
« on: October 23, 2019, 08:06:14 AM »
${y}''-{2y}'-3y=16cosh(x) r^2-2r-3=16cosh(x)\\ r^2-2r-3=16(\frac{e^{x}+e^{-x}}{2})=8e^{x}+8e^{-x}\\ (a). Homogeneous part:\\ let\,r^2-2r-3=0\\ =(r+1)(r-3),r_{1}=-1,r_{2}=3\\ y_{c}=C_{1}e^{-x}+C_{2}e^{3x}\\ Next\,we\,solve \, {y}''-{2y}'-3y=8e^{x}\\ let y_{1}=Ae^{x}\\ then\,{y}'=Ae^{x},{y}''=Ae^{x}\\ {y}''-{2y}'-3y=Ae^{x}-2(Ae^{x})-3(Ae^{x})\\ (A-2A-3A)e^{x}=8e^{x}\\ A=-2,y_{1}=-2e^{x}\\ now\,let\,{y}''-{2y}'-3y=8e^{-x}\\ let\, y_{p2}=Bxe^{-x}\,Since Be^{-x}=8e^{-x}\\ then\,{y}'=Be^{-x}-Bxe^{-x},{y}''=-Be^{-x}-Be^{-x}+Bxe^{-x}\\ {y}''-{2y}'-3y=-Be^{-x}-Be^{-x}+Bxe^{-x}-2(Be^{-x}-Bxe^{-x})-3Bxe^{-x}=8e^{-x}\\ B=-2,y_{p2}=-2xe^{-x}\\ y(x)=y_{c}+y_{p1}+y_{p2}=C_{1}e^{-x}+C_{2}e^{3x}-2e^{x}-2xe^{-x}\\ (b)y(x)=C_{1}e^{-x}+C_{2}e^{3x}-2e^{x}-2xe^{-x}\\ y(x)'=-C_{1}e^{-x}+3C_{2}e^{3x}-2e^{x}-2e^{-x}+2xe^{-x}\\ let\,y(0)'=0\\ -C_{1}+3C_{2}-2-2+0=0,-C_{1}+3C_{2}=4\\ let\,y(0)=0\\ C_{1}+C_{2}-2=0,C_{1}+C_{2}=2\\ C_{2}=\frac{2}{3},C_{1}=\frac{1}{2}\\ y=\frac{1}{2}e^{-x}+\frac{2}{3}e^{3x}-2e^{x}-2xe^{-x}$

4
##### Term Test 1 / Re: Problem 4 (noon)
« on: October 23, 2019, 06:48:44 AM »
$y''-8y'+25y=18e^{4x}+104cos(3x)\\ First \,,solve \,the\, homogenous \,eqution:\\ r^{2}-8r+25=0\\ r^{2}-8r+16=-9\\ (r-4)^{2}=-9,r-4=9i\\ r_{1}=3i+4\\ r_{2}=-3i+4\\ solution \,to \,the\, homogenous\, eqution \,is\\ y_{c}(x)=C_{1}e^{4x}cos{3x}+C_{2}e^{4x}sin{3x} now\,we\,are\,going\,to\,solve\,y_{p}(x)\\ y''-8y'+25y=18e^{4x}\\ y_{p}(x)=Ae^{4x}\\ y'=4Ae^{4x}\\ y'=16Ae^{4x}\,,then y''-8y'+25y=16Ae^{4x}-8(4Ae^{4x})+25(Ae^{4x})\\ A=2\\ Y_{P}(x)=2e^{4x}\\ and\,let's\,find\,y_{p_{2}}(x),\\ y''-8y'+25y=104cos(3x),\\ let \,y_{P}=Acos(3x)+Bsin(3x)\\ y'_{p}=-3Asin(3x)+3Bcos(3x)\\ y''_{p}=-9Acos(3x)-9Bsin(3x)\\ y''-8y'+25y=-9Acos(3x)-9Bsin(3x)-8(-3Asin(3x)+3Bcos(3x))+25(-3Asin(3x)+3Bcos(3x)\\ A=2,B=-3\,y_{P}=2cos(3x)-3sin(3x) y=y_{c}+y_{p}=C_{1}e^{4x}cos{3x}+C_{2}e^{4x}sin{3x}+2e^{4x}+2cos(3x)-3sin(3x)\\ let\,y(0)=0,y'(0)=0,\\ C_{1}=-4,C_{2}=\frac{1}{3}\\ So\,,the\,final\,solution\,is\,-4e^{4x}cos{3x}+\frac{1}{3}e^{4x}sin{3x}+2e^{4x}+2cos(3x)-3sin(3x)$

Correctly found particular and general solution, (b) incorrect V.I.

5
##### Quiz-4 / TUT 0202 QUIZ4
« on: October 18, 2019, 07:50:58 PM »
$The\,Chaacteristic\,eqution\,is:\\ 4r^{2}+12r+9=0\\ with\,roots\\ r_{1,2}=\frac{-12\pm \sqrt{144-16*9}}{8}\\ =\frac{-12\pm 0}{8}\\ =-\frac{12}{8}=-\frac{3}{2}\\ Result:y(t)=c_{1}e^{-\frac{3t}{2}}+c_{2}te^{-\frac{3}{2}}$

6
##### Quiz-3 / TUT 0202 QUIZ3
« on: October 11, 2019, 02:00:01 PM »
$we\,asume\,that\,y=e^{rt},and\,then\,it\,follows\,that\,r\,must\,be\,a\,root\,of\,characteristic\\ 2r^{2}-3r+1=(2r-1)(r-1)=0\\ Hence\\ \begin{Bmatrix} r_{1=\frac{1}{2}}\\r_{2}=1 \end{Bmatrix}\\ Since\,the\, general \,solution\, has \,the\,form\,of\\ y=c_{1}e^{r_{1}t}+c_{2}e^{r{2t}}\\ therefore,\,the\,general\,solution\,of\,the\,given\,differential\,eqution\,is\\ y=c_{1}e^{\frac{1}{2}t}+c_{2}e^{t}$

7
##### Quiz-2 / 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)$

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##### Quiz-1 / TUT 0202 QUIZ1
« on: September 27, 2019, 03:08:09 PM »
${y}'+y^{2}sin(x)= 0\\\frac{dy}{dx}= -y^{2}sin(x)\\dy=-y^{2}sin(x)dx\\-\int \frac{1}{y^{2}}dy=\int sin(x)dx\\-\frac{1}{y}=-cos(x)+C\\y=\frac{1}{cos(x)+C}$

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