Math 110 Quiz 1 || with 100% Correct Answers.

ay' + by = 0, integrating factor correct answers µ = exp(∫(b(t)/a(t))dt) ay' + by = h, variation of parameters correct answers Guess y_p=g(x)f(x), where f(x) is the homoegenous solution, and plug into l(y)=ay' + by = h. Result: g' = h(x)/[a(x)f(x)]. We have to INTEGRATE g' and MULTIPLY it by f. ay' + by = h, general solution correct answers homogeneous solution + particular solution Analytic definition correct answers If f is a smooth function, then f is analytic at x₀∈‖R iff ∃ε 0 s.t. 1) ∃(x₀-ε, x₀+ε) the Taylor series of f(x) converges 2)f = T_x₀ (Taylor expansion around x₀) at (x₀-ε, x₀+ε) Taylor series correct answers T_f, x₀ (x) = ∑_k fⁿ(x₀)/k! (x-x₀)^k, defined for all x in ‖R s.t. RHS converges ay' + by = h, Undetermined coefficients correct answers Thm: If a, b, h are analytic at x₀=0, a(0)≠0 - ∃ for each y₀∈‖R a unique solution y to ay' + by = h, y(0) = y₀ -plug in y(x) = ∑ₙ αₙxⁿ (can't go past biggest order you have multiplying a constant) ay′′+by′+cy = 0, Wronskian correct answers For y₁, y₂, W[y₁, y₂] = y₁y₂' - y₁'y₂. (Usually y1 and y2 are the two functions in the kernel.) If y₁, y₂, are linearly DEPENDENT, then W[y₁, y₂] =0. one point where it is not zero = INDEPENDENT. The matrix that the Wronskian is the determinant of is: [y₁ y₂ .... yₙ] [y₁' y₂' ... yₙ'] . . . [y₁ⁿ y₂ⁿ ... yₙⁿ] Random Theorems for 2nd Order ODE correct answers For l(y) =ay′′+by′+cy, the initial value problem is l(y) = 0 with initial conditions y(x0) =y0, y′(x0) =y1. We have seen that •Theorem: Solutions exist and every solution may be written in the form y=y0f1+y1f2, where f1 satisfies f1(0) = 1,f′1(0) = 0 and f2 satisfies f2(0) = 0, f′2(0) = 1. -For ay′′+by′+cy =0, if y solves the differential equation with x0 = 0, y(0)=y0, y'(0)=y1, the y = y0f1 + y1f2, provided that f1(0)=1, f1'(0)=0, f2(0)=0, f2'(0)=1. Linear independence correct answers Wronskian = 0 = DEPENDENT.