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Tutor profile: Stephen T.

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Stephen T.
Experienced college and high school teacher excited to work for TutorMe
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Questions

Subject: Calculus

TutorMe
Question:

Find the derivative of $$f(x) = e^{x^2 - 1}$$.

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Stephen T.
Answer:

By the chain rule, $$f'(x) = e^{x^2 - 1}\frac{d}{dx}[x^2 - 1] = e^{x^2 - 1}2x$$. So the answer is $$e^{x^2 - 1}2x$$.

Subject: Algebra

TutorMe
Question:

Factor $$81y^2 - 16x^4$$.

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Stephen T.
Answer:

We know that, if $$A$$ and $$B$$ are any numbers or expressions, $$A^2 - B^2 = (A+B)(A-B)$$. Also, we know that $$81y^2 = 9^2y^2 = (9y)^2$$ and $$16x^4 = 4^2 (x^2)^2 = (4x^2)^2$$. Therefore, $$81y^2 - 16x^4 = (9y + 4x^2)(9y - 4x^2)$$.

Subject: Differential Equations

TutorMe
Question:

Find the solution to the differential equation $$y''+y=0$$ with $$y(0)=2$$ and $$y' (0) = 0$$.

Inactive
Stephen T.
Answer:

We know that if $$\omega$$ is any real number, then the general solution to the equation $$y'' + \omega^2 y=0$$ is $$y(t) = A\cos (\omega t) + B\sin (\omega t)$$, where $$A$$ and $$B$$ are arbitrary constants. In the equation $$y''+y=0$$, we have $$\omega = 1$$, so the general solution is $$y(t) = A\cos ( t) + B\sin ( t)$$. The conditions $$y(0)=2$$ and $$y' (0) = 0$$ therefore become $$A\cos ( 0) + B\sin ( 0) = 2$$, which means $$A=2$$, and $$-A\sin ( 0) + B\cos ( 0)=0,$$ which means $$B=0$$. Therefore, the solution is $$y(t) = 2\cos(t)$$.

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