# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-5 => Topic started by: Aiting Zhang on November 01, 2019, 12:07:28 AM

Title: LEC0101 Quiz5
Post by: Aiting Zhang on November 01, 2019, 12:07:28 AM
$$\mbox{Find a particular solution of the given non homogeneous equation.}$$
$$t^2{y}''+7t{y}'+5y=t, \quad t>0; \quad y_{1}(t)=t^{-1}$$
$$\mbox{Let} \ y=v(t)y_{1}(t)$$
$$\mbox{Then} \ {y}'={v}'t^{-1}-vt^{-2}$$
$${y}''={v}''t^{-1}-{v}'t^{-2}-{v}'t^{-2}+2vt^{-3}$$
$$\mbox{Plug y, y' and y'' in the given non homogeneous equation}$$
$$\mbox{Then} \ t^{2}({v}''t^{-1}-{v}'t^{-2}-{v}'t^{-2}+2vt^{-3})+7t({v}'t^{-1}-vt^{-2})+5vt^{-1}=t{v}''+5{v}'$$
$${v}''+5{v}'t^{-1}=1$$
$$\mbox{Let} \ r={v}', \ {r}'={v}''$$
$${r}'+5{r}'t^{-1}=1$$
$$p(t)=5t^{-1}$$
$$\mu=e^{\int\frac{5}{t} dt}=t^{5}$$
$$t^{5}{r}'+5t^{4}r=t^{5}$$
$$r=\frac{1}{6}t+c_{1}t^{-5}={v}'$$
$$\mbox{Then} \ v=\int\frac{1}{6}t+c_{1}t^{-5}dt$$
$$v=\frac{1}{12}t^2-\frac{1}{4}c_{1}t^{-4}+c_{2}$$
$$\mbox{Thus,} \ y=\frac{1}{12}t^{2}t^{-1}-\frac{1}{4}c_{1}t^{-4}t^{-1}+c_{2}t^{-1}$$
$$y=\frac{1}{12}t-\frac{c_{1}}{4}t^{-5}+c_{2}t^{-1}$$
$$\mbox{Since} \ \frac{-c_{1}}{4} \mbox{is a constant. Therefore, the particular solution is } \ y=\frac{1}{12}t+c_{1}t^{-5}+c_{2}t^{-1}$$