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Tutor profile: Usama M.

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Usama M.
Mathematics Tutor
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Questions

Subject: Linear Algebra

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Question:

Fix an m × n matrix A. Then, let T : M l m → M l n, with T(B) = BA

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Usama M.
Answer:

This is a linear transformation. Let’s check the properties: (1) T(B + C) = T(B) + T(C): By definition, we have that T(B + C) = (B + C)A = BA + CA since matrix multiplication distributes. Also, we have that T(B) + T(C) = BA + CA by definition. Thus, we see that T(B + C) = T(B) + T(C), so this property holds. (2) T(dB) = dT(B): By definition, T(dB) = (dB)A = dBA while dT(B) = dBA Therefore, T(dB) = dT(B), so this property holds as well.

Subject: Applied Mathematics

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Question:

For a function y(x), we denote y 0 = dxdy the derivative with respect to x. Given the ordinary differential equation y ' + y = e( −x) with initial condition y(0) = 5

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Usama M.
Answer:

y(x) = (x + 5)e −x

Subject: Partial Differential Equations

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Question:

Consider the second-order equation in which the derivatives of second-order all occur linearly, with coefficients only depending on the independent variables: a(x, y)uxx + b(x, y)uxy + c(x, y)uyy = d(x, y, u, ux, uy).

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Usama M.
Answer:

The characteristic equation is dy dx = b ± √b2 − 4ac/2a b2 − 4ac > 0 ⇒ two characteristics, and (6.1) is called hyperbolic; b2 − 4ac = 0 ⇒ one characteristic, and (6.1) is called parabolic; b2 − 4ac < 0 ⇒ no characteristics, and (6.1) is called elliptic.

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