Overview

In this lesson, we work through the process of solving a Cauchy-Euler differential equation. These equations appear frequently in undergraduate differential equations courses and provide an excellent introduction to solving variable-coefficient ordinary differential equations using analytical techniques.

As you work through the examples, pay close attention to each algebraic step and the reasoning behind the substitutions. Understanding why each step works is just as important as obtaining the final solution.

P.L.E.M. Academy Research Position

This lesson is part of the P.L.E.M. Academy Research Position and serves as a collaborative textbook editing and research exercise.

Research students are encouraged to:

Carefully review each image and worked example.
Verify every mathematical step for accuracy.
Look for typographical, grammatical, or mathematical errors.
Suggest improvements that make the explanations clearer for future students.
Participate in the ongoing development of the P.L.E.M. Academy textbook series.

These collaborative editing sessions help produce higher-quality educational resources while giving students experience reviewing technical and scientific writing.

Discussion Questions

As you review this lesson, consider the following:

Why is the Cauchy-Euler equation considered a variable-coefficient differential equation?
When is the substitution method appropriate?
How does the characteristic equation arise from the assumed solution?
How would repeated or complex roots change the final solution?
Can you verify every algebraic manipulation independently?

Share your observations, corrections, and questions in the discussion.

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admin

21 hours ago

This is a sample of what the code looks like and what it looks like parsed. You as a student will type your responses in Microsoft Word, have ChatGPT or any A.I. retype in LaTeX under Jetpack rules, and then post here.

$\Large \begin{aligned} \textbf{By Cauchy\!-\!Euler Method: Let } y=x^r. \end{aligned}$

$\Large \begin{aligned} y&=x^r,\\[1em] y'&=rx^{r-1},\\[1em] y''&=r(r-1)x^{r-2},\\[1em] y'''&=r(r-1)(r-2)x^{r-3}. \end{aligned}$

$\Large x^3y'''+4x^2y''+10xy'-10y=0$

$\Large \begin{aligned} \Rightarrow\;& x^3\!\left[r(r-1)(r-2)x^{r-3}\right] +4x^2\!\left[r(r-1)x^{r-2}\right] +10x\!\left[rx^{r-1}\right] -10\!\left[x^r\right]=0 \end{aligned}$

$\Large \begin{aligned} \Rightarrow\;& \left[r(r-1)(r-2)x^{\,r-3+3}\right] +4\left[r(r-1)x^{\,r-2+2}\right] +10\left[rx^{\,r-1+1}\right] -10\left[x^r\right]=0 \end{aligned}$

$\Large \begin{aligned} \Rightarrow\;& \left[r(r-1)(r-2)x^r\right] +4\left[r(r-1)x^r\right] +10\left[rx^r\right] -10\left[x^r\right]=0 \end{aligned}$

$\Large \begin{aligned} \Rightarrow\;& x^r\left(r(r-1)(r-2)+4r(r-1)+10r-10\right)=0 \end{aligned}$

$\Large \begin{aligned} \Rightarrow\;& x^r>0,\\[0.75em] &r(r-1)(r-2)+4r(r-1)+10r-10=0 \end{aligned}$

$\Large \begin{aligned} \Rightarrow\;& r(r^2-3r+2)+4r^2-4r+10r-10=0 \end{aligned}$

$\Large \begin{aligned} \Rightarrow\;& r^3-3r^2+2r+4r^2-4r+10r-10=0\\[0.75em] \Rightarrow\;& r^3+r^2+8r-10=0 \end{aligned}$

$\Large \left\{\pm1\right\};\; \left\{\pm1,\pm2,\pm5,\pm10\right\}$

$\Large \begin{aligned} r^3+r^2+8r-10 =(\pm1)^3+(\pm1)^2+8(\pm1)-10=0\ldots \end{aligned}$

$\Large \begin{aligned} \text{Clearly,}\qquad r=1 \qquad\Rightarrow\qquad (r-1)=0. \end{aligned}$

Last edited 21 hours ago by admin

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