# Toronto Math Forum

## MAT244--2018F => MAT244--Tests => Quiz-4 => Topic started by: Victor Ivrii on October 26, 2018, 05:40:42 PM

Title: Q4 TUT0101
Post by: Victor Ivrii on October 26, 2018, 05:40:42 PM
Verify that the given functions $y_1$ and $y_2$ satisfy the corresponding homogeneous equation; then find a particular solution of the given inhomogeneous equation.
\begin{gather*}
\end{gather*}
Title: Re: Q4 TUT0101
Post by: Monika Dydynski on October 26, 2018, 05:43:58 PM
Verify that the given functions $y_1$ and $y_2$ satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation.

$$\begin{gather*} t^2y''-2y=3t^2-1,\qquad t>0\tag{1};\\ y_1(t)=t^2,\quad y_2(t)=t^{-1}. \end{gather*}$$

1. Verify that $y_1(t)=t^{2}$ and $y_2(t)=t^{-1}$ satisfy the corresponding homogeneous equation,

$$t^{2}y''-2y=0.\tag{2}$$

$$\cases{y_1(t)=t^{2}\\y'_1(t)=2t\\y''_1(t)=2}$$

Plugging $y_1$, $y'_1$, and $y''_1$ into $(2)$,

$$t^{2}(2)-2(t^{2})=0 \Rightarrow 2t^{2}-2t^{2}=0$$

$y_1$ satisfies $(2)$ $\Rightarrow$ $y_1$ is a solution to the corresponding homogeneous equation.

Similarily,

$$\cases{y_2(t)=t^{-1}\\y'_2(t)=-t^{-2}\\y''_2(t)=2t^{-3}}$$

Plugging $y_2$, $y'_2$, and $y''_2$ into $(2)$,

$$t^{2}\left(\frac{2}{t^{3}}\right)-2\left(\frac{1}{t}\right)=0 \Rightarrow \frac{2}{t}-\frac{2}{t}=0$$

$y_2$ satisfies $(2)$ $\Rightarrow$ $y_2$ is a solution to the corresponding homogeneous equation.

2. Find a particular solution of the given nonhomogeneous equation,

$$t^{2}y''-2y=3t^{2}-1, t>0.\tag{1}$$

Dividing $(1)$ by $t^{2}$, we have

$$y''-\frac{2}{t^{2}}y=3-\frac{1}{t^{2}}$$

Note that $p(t)=0$, $q(t)=-\frac{2}{t^{2}}$, and $g(t)=3-\frac{1}{t^{2}}$ are continuous on $(0, +\infty)$

Computing the Wronskian,

$$W(t^{2}, t^{-1})(t)=\begin{array}{|c c|}t^{2}& t^{-1} \\ 2t & -t^{-2}\end{array}=-1-2=-3\ne0,$$

we verify that $y_1(t)=t^{2}$ and $y_2(t)=t^{-1}$ form a fundamental set of solutions $\forall t>0$

Calculating the parameters $u_1$ and $u_2$, we get

$$u_1=-\int{\frac{y_2(t)g(t)}{W(y_1,y_2)(t)}}dt=-\int{\frac{t^{-1}(3-t^{-2})}{-3}}dt=\frac{1}{3}\int{\frac{3}{t}-\frac{1}{t^{3}}}dt=\ln{t}+\frac{1}{6t^{2}}+c_1$$

$$u_2=\int{\frac{y_1(t)g(t)}{W(y_1,y_2)(t)}}dt=\int{\frac{t^{2}(3-t^{-2})}{-3}}dt=-\frac{1}{3}\int{3t^{2}-1}dt=-\frac{t^{3}}{3}+\frac{t}{3}+c_2$$

The particular solution is

$$y_p(t)=t^{2}\left(\ln{t}+\frac{1}{6t^{2}}+c_1\right)+\frac{1}{t}\left(-\frac{t^{3}}{3}+\frac{t}{3}+c_2\right)=t^{2}\ln{t}+\frac{1}{6}+c_1t^{2}-\frac{t^{2}}{3}+\frac{1}{3}+\frac{c_2}{t}=t^{2}\ln{t}+\frac{1}{2}+c_1t^{2}+\frac{c_2}{t}$$

$$y_p(t)=t^{2}\ln{t}+\frac{1}{2}.$$
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9932705760002136, "perplexity": 674.2752042059849}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585025.23/warc/CC-MAIN-20211016200444-20211016230444-00098.warc.gz"}