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# Tutor profile: Sergio M.

Sergio M.
Been teaching college-level math for several years

## Questions

### Subject:Basic Math

TutorMe
Question:

Is it possible that Anna will be twice as old as Ben in two years, but she was born two years before him?

Sergio M.

One option would be to try and guess, which is not too hard in this case. The systematic way would be: denote by $A$ Anna's current age, and by $B$ Ben's current age. The first bit of information means that $$A + 2 = 2(B+2),$$ while the second one means that $$A = B + 2.$$ But then, $A + 2 = B+4,$ from the second equation, so the first one reads $$B + 4 = 2(B+2) = 2B + 4,$$ which forces $B=0$, so Ben is right now being born, and Anna's age is $A = 0 + 2 = 2.$ So it is indeed possible.

### Subject:Calculus

TutorMe
Question:

A mountain has a circular base of radius $1$ km; assume the center of the circle coincides with the origin in the $xy$-plane. The elevation at point $(x,y)$ is $h(x,y) = 1-x^2-y^2$. Find the average elevation of the mountain.

Sergio M.

By definition of average of a function, the average elevation should be $$\frac{1}{A(D)} \int{D} h(x,y) dx dy$$ where $D$ is the circular base of the mountain (a disk of radius $1$) and $A(D)$ is its area. Thus, we need to calculate the double integral $$\frac{1}{\pi} \int_D (1-x^2-y^2)dx dy,$$ which is easier done in polar coordinates: $x = r \cos\theta,$ $y= r\sin\theta,$ $r^2 = x^2 + y^2$, and the area element $dxdy = r dr d\theta,$ so we have $$\frac{1}{\pi} \int_0^{2\pi}\int_0^1 (1-r^2) r dr d\theta = 2 \int_0^1 (r-r^3) dr = 1/2.$$

### Subject:Differential Equations

TutorMe
Question:

Find all solutions to the equation $$x^2 y'' - 3x y' + y = 0.$$

Sergio M.

Notice that the coefficients of the equation have decreasing exponents, and differentiation of power functions decreases the exponent. One is tempted to look for a solution of the form $y = x^m.$ This is indeed called a Cauchy-Euler equation. If one tries $y= x^m$ into the equation, one gets, for the left-hand side: $$x^2 m(m-1)x^{m-2} - 3x m x^{m-1} + x^m = m(m-1) x^m - 3m x^m + x^m = [m(m-1) - 3m + 1]x^m$$ so we should find $m$ that will make this zero, which means solving $$m^2 - 4m + 1 = 0,$$ giving us $m = 2-\sqrt{3}$ and $m = 2+\sqrt{3}$. Thus, $$y_1 = x^{2-\sqrt{3}}, \qquad y_2 = x^{2+\sqrt{3}}$$ are two solutions of the given equation. We can compute the Wronskian of this two functions, and we will get $$W(y_1,y_2) = (2+\sqrt{3})x^{2+\sqrt{3} + 2 -\sqrt{3}} - (2 - \sqrt{3})x^{2+\sqrt{3} + 2 -\sqrt{3}},$$ which is different from zero. So the two solutions are linearly independent. Since the equation is of second order, we have found all the solutions: $$y = C_1 x^{2-\sqrt{3}} + C_2 x^{2+\sqrt{3}}.$$

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