Author Topic: LEC 5201 QUIZ5  (Read 548 times)

Kunpeng Liu

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LEC 5201 QUIZ5
« on: November 01, 2019, 11:08:13 AM »
$$Question:Find\, \, a \, \, \, \,particular \, \, solution \, \, of\, \,  the\, \,  given\, \,  nonhomogeneous\, \,  equation.\, \, \, \,  \\\\t^2{y}''+7t{y}'+5y=t,\, \, \, \, \,  t> 0; \, \, \, \, \, \, \, \, \, y(1)=t^{-1}\\\\Let\, \, y=V(t)t^{-1}\, \, \, \, \, \, \, \, {y}'={v}'t^{-1}+v(-1)t^{-2}\, \, \, \, \, \, \, \, \, \, \, \\\\ {y}''={v}''t^{-1}+{v}'(-1)t^{-2}+(-1){v}'t^{-2}+(-1)v(-2)t^{-3}\\\\Substituted\, \, \, y\, \,\, \, \,  {y}'\, \, \, {y}''\, \, \, \, in\, \, \, \, t^2{y}''+7t{y}'+5y=t:{V}''+5t^{-1}{v}'=1\\\\let\, \, \, \,  {v}''={r}'\, \, \, {v}'=r\, \, \, \, then\, \, {r}'+5t^{-1}r=1\\\\\mu =e^{\int (5/t)dt}=t^5\\\\t^5{r}'+5t^4r=t^5\, \, \, \, {(t^5r)}'=t^5\, \, \, \, t^5=t^6(1/6)+c\\\\r=(1/6)t+ct^{-5}={v}'\\\\V=(1/12)t^2+(c/-4)t^{-4}+C2\\Y=(1/12)t+C1t^{-5}+C2t^{-1}$$
« Last Edit: November 01, 2019, 11:11:24 AM by Kunpeng Liu »