Science:Math Exam Resources/Courses/MATH101/April 2009/Question 01 (h)

MATH101 April 2009

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Question 01 (h)
For what values of $r$ does the function $y=x^{r}$ satisfy the following differential equation (for $x>0$)? $\displaystyle x^{2}y''+4xy'+2y=0$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Try plugging in the values of $y,y',y''$ into the differential equation.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. When $y=x^{r}$, we know that $y'=rx^{r-1}$ and $y''=r(r-1)x^{r-2}$. Hence, we have ${\begin{aligned}0&=x^{2}y''+4xy'+2y\\&=x^{2}(r(r-1)x^{r-2})+4x(rx^{r-1})+2x^{r}\\&=(r^{2}-r)x^{r}+4rx^{r}+2x^{r}\\&=(r^{2}+3r+2)x^{r}\end{aligned}}$ Now, since $x>0$ always, we must have that the coefficient of $x^{r}$ above has to be 0. Thus, we solve $\displaystyle 0=r^{2}+3r+2=(r+1)(r+2)$ and so $r=-1$ and $r=-2$ completing the question.

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