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Messages - Kunpeng Liu

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1
Quiz-5 / LEC 5201 QUIZ5
« on: November 01, 2019, 11:08:13 AM »
$$Question:Find\, \, a \, \, \, \,particular \, \, solution \, \, of\, \,  the\, \,  given\, \,  nonhomogeneous\, \,  equation.\, \, \, \,  \\\\t^2{y}''+7t{y}'+5y=t,\, \, \, \, \,  t> 0; \, \, \, \, \, \, \, \, \, y(1)=t^{-1}\\\\Let\, \, y=V(t)t^{-1}\, \, \, \, \, \, \, \, {y}'={v}'t^{-1}+v(-1)t^{-2}\, \, \, \, \, \, \, \, \, \, \, \\\\ {y}''={v}''t^{-1}+{v}'(-1)t^{-2}+(-1){v}'t^{-2}+(-1)v(-2)t^{-3}\\\\Substituted\, \, \, y\, \,\, \, \,  {y}'\, \, \, {y}''\, \, \, \, in\, \, \, \, t^2{y}''+7t{y}'+5y=t:{V}''+5t^{-1}{v}'=1\\\\let\, \, \, \,  {v}''={r}'\, \, \, {v}'=r\, \, \, \, then\, \, {r}'+5t^{-1}r=1\\\\\mu =e^{\int (5/t)dt}=t^5\\\\t^5{r}'+5t^4r=t^5\, \, \, \, {(t^5r)}'=t^5\, \, \, \, t^5=t^6(1/6)+c\\\\r=(1/6)t+ct^{-5}={v}'\\\\V=(1/12)t^2+(c/-4)t^{-4}+C2\\Y=(1/12)t+C1t^{-5}+C2t^{-1}$$

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Term Test 1 / Re: Problem 4 (afternoon)
« on: October 23, 2019, 10:24:04 AM »
Hello Mengyuan Wang´╝îyour question should be Problem 4(afternoon) rather than Problem 4(morning)

3
Term Test 1 / Re: Problem 4 (morning)
« on: October 23, 2019, 09:35:47 AM »
$$Let \, \, \, y=e^{rt}\, \,\, \,  {y}'=re^{rt}\, \, \, {y}''=r^{2}e^{rt}\\\\Then\, \, \, \, \, \, r^{2}-6r+25=0\, \, \, \, \, \, \, \\\\\\\\\\r1=\frac{6+\sqrt{36-4\cdot 25\cdot 1}}{2}=3+4i\, \, \, \, \, \, \, r2=\frac{6-\sqrt{36-4\cdot 25\cdot 1}}{2}=3-4i\\\\Yc(x)=C1e^{3x}cos4x+C2e^{3x}sin4x\\\\\\\\\\Let\, \, Yp(x)=Ae^{3x}\, \, \,\, \, \,  {y}'=3Ae^{3x}\, \, \,\, \,  {y}''=9Ae^{3x}\\\\\\\\\\9Ae^{3x}-18Ae^{3x}+25Ae^{3x}=16e^{3x}\, \, \, \, \, \, 16Ae^{3x}=16e^{3x}\, \, \, \, \, 16A=16\, \,\, \, \, \,  A=1\\\\\\\\\\Yp(x)=e^{3x}\\\\\\\\\\let \, \, \, \, Yq(x)=Asinx+Bcosx\, \, ,\, \, {y}'=Acosx-Bsinx\, \, ,\, \, {y}''=-Asinx-Bcosx\\\\\\\\\\-Asinx-Bcosx-6Acosx+6Bsinx+25Asinx+25Bcosx=102sinx\\\\\\\\\\(24A+6B)sinx+(24B-6A)cosx=102sinx\, \, \, \, \, \, 24A+6B=102\, \, \, \, 24B-6A=0\\\\A=4\, \, \, \, B=1\\\\\\\\\\Yq(x)=4sinx+cosx\\\\\\\\\\Y(x)=C1e^{3x}cos4x+C2e^{3x}sin4x+e^{3x}+4sinx+cosx\\\\y(0)=0\\\\\\because y(0)=C1+2=0\, \, \, \, C1=-2\\\\\\\\\\\because\, \, \, \, \,  {y}'(0)=0\, \, \, \, {y}'(0)=3C1-0+0+4C2+3+4=0\, \, \, \, C2=-\frac{1}{4}\\\\\\\\\\Y(x)=-2e^{3x}cos4x-\frac{1}{4}e^{3x}sin4x+e^{3x}+4sinx+cosx$$

4
Quiz-4 / TUT 0702 QUIZ4
« on: October 18, 2019, 02:00:00 PM »
$$Question:\, \, Find \, \, the\, \,  general \, \, solution\, \,  of\, \,  the\, \,  given\, \,  differential\, \,  equation:{y}''+2{y}'+2y=0\\\\\\To \, \, begin\, \,  with,let\, \, y=e^{rt},{y}'=re^{rt},{y}''=r^2e^{rt}\\\\\\\\Then, r^2+2r+2=0, \, \, r1=\frac{-2+2i}{2}=-1+i,\, \, r2=\frac{-2-2i}{2}=-1-i\\\\\\\\Substitute \, \, \lambda =-1\, \, and \, \, \mu =1\, \, in \, \,\, \,  y=C1e^{\lambda t}cos(\mu t)+C2e^{\lambda t}sin(\mu t):\\\\\\\\y=C1e^{-t}cost(t)+C2e^{-t}sin(t)$$

5
Quiz-3 / TUT 0702 QUIZ3
« on: October 11, 2019, 02:00:02 PM »
$$Question:Find\, \, the\, \, Wronskian\, \, of\, \, two\, \, solutions\, \, of\, \, the\, \, given\, \, differential\, \, equation\, \, without\, \, solving\, \, the\, \, equation.\\\\\\t^{2}{y}''-t(t+2){y}'+(t+2)y=0\\\\To\, \, begin\, \, with,\, divide\, \, both\, \,  sides \, \, of \, \, the\, \,  equation\, \,  by\, \,  t^{2}\\\\y{}''-\frac{t+2}{t}y{}'+\frac{t+2}{t^2}y=0\\\\Let\, \, \, \,  P(t)=-\frac{t+2}{t}\\\\Then \, \, \, \, W=Ce^{-\int-\frac{t+2}{t}dt }=Ce^t\cdot e^{2lnt}=Ct^2e^t$$

6
Quiz-2 / TUT 0702 QUIZ2
« on: October 04, 2019, 03:03:04 PM »
$$ Question:Show\, \, that\, \, the\, \, given\, \, equation\, \, is\, \, not\, \, exact\, \, but\, \, becomes \, \, exact\, \,  when\, \,  multiplied\, \,  by\, \,  the \, \, given\, \,  integrating\, \,  factor.\, \, Then\, \, solve \, \,the \, \, equation.\\x^{2}y^{3}+x(1+y^{2}){y}'=0\, \, \, \, \, \, \mu (x,y)=ye^x\\\\Let\, \, M=x^2y^3,\, \, N=x(1+y^2)\\\\My=3x^2y^2,\, \, Nx=1+2xy\, \, \, \, \because My\neq Nx\, \, \, \therefore Not\, \, \, exact\\\\Both\, \, sides\, \, multiplied\, \, \mu :\\\\x+(\frac{1}{y^3}+\frac{1}{y}){y}'=0\, \, \, \, \, \, Let\, \, M=x\, \, \, \, N=\frac{1}{y^3}+\frac{1}{y}\\\\\because My=Nx=0\, \, \therefore Exact\\\\\phi x=\int xdx=\frac{1}{2}x^2+h(y)\, \, \, \, \, \, \, \, \, \, \phi y={h}'(y)=y^{-3}+y^{-1}\\\\h(y)=-\frac{1}{2}y^{-2}-ln\left | y \right |\\\\\therefore \frac{1}{2}x^{2}-\frac{1}{2}y^{-2}-ln\left | y \right |=C$$

7
Quiz-2 / TUT0702 QUIZ2
« on: October 04, 2019, 02:18:56 PM »
 $$Question:Show\, \, that\, \, the\, \, given\, \, equation\, \, is\, \, not\, \, exact\, \, but\, \, becomes \, \, exact\, \,  when\, \,  multiplied\, \,  by\, \,  the \, \, given\, \,  integrating\, \,  factor.\, \, Then\, \, solve \, \,the \, \, equation.\\x^{2}y^{3}+x(1+y^{2}){y}'=0\, \, \, \, \, \, \mu (x,y)=ye^x\\\\Let\, \, M=x^2y^3,\, \, N=x(1+y^2)\\\\My=3x^2y^2,\, \, Nx=1+2xy\, \, \, \, \because My\neq Nx\, \, \, \therefore Not\, \, \, exact\\\\Both\, \, sides\, \, multiplied\, \, \mu :\\\\x+(\frac{1}{y^3}+\frac{1}{y}){y}'=0\, \, \, \, \, \, Let\, \, M=x\, \, \, \, N=\frac{1}{y^3}+\frac{1}{y}\\\\\because My=Nx=0\, \, \therefore Exact\\\\\phi x=\int xdx=\frac{1}{2}x^2+h(y)\, \, \, \, \, \, \, \, \, \, \phi y={h}'(y)=y^{-3}+y^{-1}\\\\h(y)=-\frac{1}{2}y^{-2}-ln\left | y \right |\\\\\therefore \frac{1}{2}x^{2}-\frac{1}{2}y^{-2}-ln\left | y \right |=C$$

8
Quiz-1 / Re: TUT 0702 QUIZ1
« on: September 27, 2019, 02:32:21 PM »
$$
Find\, \, \, \,the\, \, \, solution\, \, of\, \,  the\, \, \,  given\, \,  initial\, \,  value\, \,  problem\, \,  in\, \,  explicit\, \,  form:{y}'=2x/(1+2y), y(2)=0\\\\\frac{dy}{dx}=\frac{2x}{1+2y}\\\\(1+2y)dy=2xdx\\\\\int (1+2y)dy=\int 2xdx\\\\(y+1/2)^{2}=x^{2}+c\\\\y+\frac{1}{2}=(x+c)^{1/2}\\\\y=(x+c)^{1/2}-\frac{1}{2}\\\\\because y(2)=0 \rightarrow 0=(4+c)^{1/2}-\frac{1}{2}\\\\therefore\, \, C=\frac{-15}{16}\\\\y=(x^{2}-\frac{15}{16})^{1/2}-\frac{1}{2}
$$

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Quiz-1 / TUT 0702 QUIZ1
« on: September 27, 2019, 02:29:41 PM »
$$
Find\, \, \, \,the\, \, \, solution\, \, of\, \,  the\, \, \,  given\, \,  initial\, \,  value\, \,  problem\, \,  in\, \,  explicit\, \,  form:{y}'=2x/(1+2y), y(2)=0\\\\\frac{dy}{dx}=\frac{2x}{1+2y}\\\\(1+2y)dy=2xdx\\\\\int (1+2y)dy=\int 2xdx\\\\(y+1/2)^{2}=x^{2}+c\\\\y+\frac{1}{2}=(x+c)^{1/2}\\\\y=(x+c)^{1/2}-\frac{1}{2}\\\\\because y(2)=0 \rightarrow 0=(4+c)^{1/2}-\frac{1}{2}\\\\therefore\, \, C=\frac{-15}{16}\\\\y=(x^{2}-\frac{15}{16})^{1/2}-\frac{1}{2}
$$

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