# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-5 => Topic started by: Qihui Huang on November 01, 2019, 02:03:28 PM

Title: LEC0101 quiz5
Post by: Qihui Huang on November 01, 2019, 02:03:28 PM
Question: find a particular solution of the given homogeneous equation:
$$t^2y''+7ty'+5y=t, t>0$$
Rewrite as $$y''+\frac{7}{t}y'+\frac{5}{t^2}y=\frac{1}{t}$$ Let $$y=v(t)y_1(t)=vt^{-1}$$ then $$y'=v't^{-1}-vt^{-2}$$ $$y''=v''t^{-1}-2v't^{-2}+2vt^{-3}$$
$$v''t^{-1}-2v't^{-2}+2vt^{-3}+7v't^{-2}-7vt^{-3}+5vt^{-3}=t^{-1}$$ $$v''t^{-1}+5t^{-2}v'=t^{-1}$$ $$v''+5t^{-1}v'=1$$
Let $$v'=r(t), v''=r'(t)$$ so we get $$r'+\frac{5}{t}r=1$$ Find the integrating factor $$u(t)=e^{\int \frac{5}{t}dt}$$ $$u(t)=t^5$$ $$t^5r'+t^5\frac{5}{t}r=t^5$$ $$\int (t^5r)'=\int t^5 dt$$ $$rt^5=\frac{1}{6}t^6+c_1$$ $$r=\frac{1}{6}t+c_1t^{-5}$$ $$v=\frac{1}{12}t^2-\frac{1}{4}c_1t^{-4}+c_2$$ $$y=\frac{1}{12}t-\frac{1}{4}c_1t^{-4}t^{-1}+c_2t^{-1}$$ $$y=\frac{1}{12}t-\frac{1}{4}c_1t^{-5}+c_2t^{-1}$$ $$y=a_1t^{-5}+a_2t^{-1}+\frac{1}{12}t$$where $a_1$ and $a_2$ are arbitrary constants