If u(x, t) is the solution of then ди = J²u It მე-2 ა ,0 < x < 1,t > 0, u(x, 0) = 1+x+ sin (πx) cos (πx), u(0, t) = 1, u(1, t) = 2, (a) u() = (b) (ਨੂੰ, ਨੂੰ) = (c) u()+½½-3² = (d) u(1, 1) = e−4 -42

Here i is the imaginary unit.a) Use the Euler formula, e^(iθ) = cos(θ) + i sin(θ), to prove that for any n, we have(cos(θ) + i sin(θ))^n = cos(nθ) + i sin(nθ).b) Suppose y(t) = e^(−2t)e^(it) solves y′′ + By′ + C = 0. What are B and C?

In implicit differentation problems it seems we treat x and y differently. For example, when we differentiate both sides of cos(x) + sin(y) = x^3 + y^3, we get -sin(x) + cos(y)y' = 3x^2 + 3y^2y'. a) Why do we get y' but but not a x'? WHy are these variables treated differently? b) How is this related to the chain rule?

1. Let s(t) = 1 -t 1+ et ca) find s' (t) (b) Which of the following equations hold true? Show why your choice is true. [Note: Only 1 equations is true]. i. s' (t) = s(t) ii. s' (t) = (s(t))² iii. s' (t) = s(t) (1 -S(t)) iv. Ś(t) = -S(-t)