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Determine all values of the constant $r$ such that the given function solves the given differential equation.

$$y(x)=x^{r}, \quad x^{2} y^{\prime \prime}+x y^{\prime}-y=0$$.

$r=\pm 1$

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in this problem, we are given a differential equation, and we want to figure out for what are this way of X solves this differential equation. So to do that, we will start by, um, finding expressions for y prime. And why double prime? So the derivative of why the axe is going to be acts times but sorry r Times X to the R minus one. That's just taking the power role. And the second derivative by double prime of X is equal. Teoh are times ar minus one times x to the AR minus two. All right, so now we have these expressions. We can go ahead and plug them into our differential equation and figure out for what are that makes our differential equation equal to zero. So we start by writing x two times Why Double Prime, which is our r minus one times x to the R minus two and plus x times y prime So x times R Times X to the R minus one and then minus Why, which is X to the R, and we want to figure out when this is equal to zero. Well, we can simplify this a bit because X squared times x to the R minus two. When we multiply things with exponents, we just add the exponents. So we get that That's just equal to extra. They are. So we get our times ar minus one x to the arm plus again X times extra ar minus one is just X to the are so plus R X to the arm minus XDR, and we want to figure out when that's equal to zero. We can factor out extra the are and rewrite the inside as our times ar minus one is R squared minus are plus our minus one, and we want to figure out when that's equal to zero while extra the are can never equal zero. So we're really just concerned about when this inside equation will equal zero minus are in our cancel. So we want to know When does R squared minus one equals zero? What is the difference of two squares? So we get this is equal Teoh. When extra, the R times ar minus one. It's far plus one equals zero. And that's true exactly when R equals one or R equals negative one. So if r equals one or negative one, that's when, um, that's when why that solves this differential