MA 262 Exam 1 Study Guide
MA 262 Exam 1 Study Guide MA 262
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This 6 page Study Guide was uploaded by Abby Frazier on Saturday September 24, 2016. The Study Guide belongs to MA 262 at Purdue University taught by Professor Ying Zhang in Fall 2016. Since its upload, it has received 26 views. For similar materials see Linear Algebra and Differential Equations in Math at Purdue University.
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Date Created: 09/24/16
MA 262 Exam 1 Study Guide: Chapter 1: Main Ideas: 1. Seperable: p(y)dy = q(x)dx à integral p(y)dy = integral q(x)dx +c st 2. 1 order linear: y’ + p(x)y = g(x) à I(x) = e^integral p(x)dx à sty)’ = Ig(x) 3. 1 order homogenous: y’ = F(x, y) à V = y/x –separable DE for V ày = Vx 4. Berinoilli n Y’ + p(x)1 – n(x)y à V = y –first order linear for V à y = 1/(V 1 – ) 5. Exact DE Mdx + Ndy = 0 à integral Mdx à integral Ndy *and don’t repeat terms 1.2 Basic Ideas: • Definitions: o Linear/nonlinear differential equation § y, y’, y”, y’” … all must have a power of 1 BUT the x terms may have non linear terms o solution § any differential function y = f(x) is a solution to a DE if it satisfies the DE • Order of a differential equation – the order of a DE is the order of the highest derivative occurring in the DE • General form of an nth order linear DE (n) (n-1) (1) o a (x0y + a (x)1 +… a n-1y + a (x)n = f(x) àif it is not in this form then the DE is non linear o only care about the y terms NOT the x terms st 1.3 The geometry of a 1 order DE • Definitions: o solution curve § the graph of any solution to the DE o equilibrium solution § any solution to f(x, y) = 0 with y = constant is an equilibrium solution to the DE dy/dx = f(x, y) • Theorem: o Existence and uniqueness theorem dy/dx = f(x,y) y(x 0 = y 0– (x0, y0) -has a unique solution in R = (x,y)|a =< x =< b |c =< y =< d where (x ,0 )0is inside the region of R, if both f(x, y) and partial f/partial y (x,y) are continuous functions in R because each family of curves has to follow the same shape and will therefor never touch § Slope field—sketch many small line segments which present the solution curve 1.4 § Definitions: o Separable DE § A first order DE is called separable if it can be written in the form p(y) dy/dx = q(x) § Theorem: o If q(y) and q(x) are continuous functions, then p(y) dy/dx = q(x) has a solution Integral p(y) dy = integral q(x) dx +c § Equations: o dT/dt = -k(T – T ) 0 § T = temperature, t = time, k = constant (negative because cooling), T = 0 room temperature (constant), dT/dt = rate of change proportional to the difference of temperature of the object and room temperature multiplied by k § T = T 0à equilibrium solution T = f(t) If T is not equal to T 0then dT/(T – T ) 0 -kdt T = T 0+ ce^-kt c >= 0 à general solution for any newton’s law of cooling 1.5 Simple Population Models § Malthusian growth model – no limit on the growth of the population dP/dt = kP where k is constant P = ce^(kt) c > 0 § Definition: o Doubling time t d § The time required for a culture to double in size t is tde time for P = 2P 0 § td = ln(2)/k § Logistic Population Model – a limit on the growth of the population dP/dt = r(1 – P/c)P -r = rate, c = capacity of population P = cP0/(P 0+ (c – P 0e^(-rt)) st 1.6 1 order linear DE § Definition: o A 1 order DE has the form a(x) dy/dx +b(x)y = c(x) OR dy/dx +p(x)y = g(x) à standard form § If a 1 order DE is in the standard form: 1. Let I(x) = e^(integral p(x)dx not equal to zero -then multiply both sides of the equation by I(x) 2. (I(x)y)’ = I(x)g(x) 3. y =( integral I(x)g(x) dx +c)/I(x) st 1.7 Application for 1 order DE § Mixing Problem Equations: o v = volume, c = concentration, r = rate, A = amount \ o v = (r 1 r )2 + c ; c = v0 o dA/dt = c r 1 1 r 2 2c r 1 1r /(2r –1r )2 + v )0 1.8 Change of Variables Lecture § Definition: o f(x,y) is a homogenous function of degree 0 if f(tx, ty) = f(x,y) for all partial values t o Berinoilli equation has the form dy/dx + p(x)y = g(x)y where n is constant st -n = 0 is a 1 order DE -n = 1 is seperable § Berinoilli equation: 1-n -n 1. Set u = y ; du/dx = 1-n(y) dx/dy 2. Substitute into equation à 1/(1-n) du/dx + p(x)u = g(x) n 3. Divide both sides by y to get du/dx by itself 4. Solve as a 1 order DE 1.9 Exact DE § Definition o Exact DE § M(x, y)dx + N(x, y)dy = 0 = M(x, y) + N(x, y)dy/dx is exact if there exists a function ø(x, y) such that ∂ø/∂x = M(x, y) and ∂ø/∂y = N(x, y) § ø(x, y) is called a potential function of the DE o An non zero function I(x, y) is an integrating factor for Mdx +Ndy = 0 IF I(x, y)M(x , y)dx +I(x, y)N(x, y)dy = 0 is exact § Theorem o Test for exactness § M(x, y)dx + N(x,y)dy = 0 is exact if and only if ∂M/∂y = ∂N/∂x § The general solution to an exact DE is ø(x, y) = c § If ∂M/∂y does NOT equal ∂N/∂x 1. Find integrating factor by steps below 2. I(x, y) = e^integral f(x)dx or I(x, y) = e^integral g(y)dy 3. Solve new M and N values o Consider M(x, y)dy + N(x, y)dy = 0 1. If (M – N /N = f(x), then I(x, y) = e^integral f(x)dx à function of x ONLY y x( 2. If (N x– M /y(= g(y), then I(x, y) = e^integral g(y)dx à function of y ONLY 3. Solve for ø(x, y) as before § Solving for DE during exam: 1. Integral M(x, y)dx = Integral N(x, y)dy = 2. Then take both integrals and add each term together once, ignoring repeating terms Ex) = 3x+4y = 3x + 123 ø(x, y) = 3x + 123 = c 1.10 nd § 2 order DE y” = F(x, y, y’) V = dy/dx dV/dx = y’’ à dV/dx = F(x, y, y’) o cases in which we can use dV/dx: 1. with y missing in F dV/dx = F(x, V) à solve for V from above equation dy/dx = V à solve for y 2. with x missing in F dV = F(y,V) dy/dx = V d y/dx = d/dx (dy/dx) = dV/dx = dV/dy/V = V dV/dy = F(y, V) Chapter 2: 2.1 Matrix Vectors • Definition o Matrix § An m x n matrix is a rectangular array of numbers arranged in m rows and n columns § Square matrix = n x n § Represented with capital letters § A = [aij]; I = ith row; j = jth column o A = B if 1. A and B are the same size 2. Aij = bij for all ij o Row vectors/column vectors § A 1 x n matrix is a ROW vector [1 3 5] 1 x 3 § A 1 x m matrix is a COLUMN vector [1 5 3] 3 x 1 o Transpose of A (A ) T § A is an m x n matrix in which we interchange the row vectors and the column vectors in the m x n matrix A A = [aij] à A = [aij] § An n x m matrix is called a square matrix with a main diagonal trace § IF A is a square matrix, then aii (1 =< I =< n) make up the main diagonal of leading diagonal of the matrix § Trace of A = tr(A) = a 11 a 22 … +a nn o Upper/lower triangular, diagonal matrix § A square matrix A is upper if aij = 0 whenever I > j § A square matrix A is lower if aij = 0 whenever I < j o Symmetric/skew/anti skew § A square matrix A is symmetric if A = A T T § A square matrix A is skew-symmetric if A = -A § A is symmetric if aij = aji § A is skew symmetric if aij = -aji and if the main diagonal is zero 2.2 Matrix Algebra § Definition: o A +/- B à where A and B are the same size § A +/- B = [aij + bij] or [aij – bij] § s-scalar: sA = [saij] § Property: 1. A + B = B + A 2. A + (B + C) = (A + B) + C 3. 1A = A (unit property) 4. s(A + B) = sA + sB 5. (s+t)A = sA +tA 6. s(tA) = (st)A = t(sA) 7. A + 0 = A *where 0 is a zero matrix with the same size as A A – A = zero matrix 0 x A = 0 § Multiplication of matric: 1. Product of matrix sizes 1 n and n x 1 a x b = a1 1 + a2 2 + … + a n n 2. Product of matrix sizes m x n and n x 1 Ab = a 11 1 a b12 2+ a b 1n n a21 1+ a 22 2…+ a b 2n n a b + a b +…+ a b m1 1 m2 2 mn n 3. Product of matrix sizes m x n and n x p C = AB = ▯ a1i + b i1 ▯ a1i + b i2 … ▯ a1i + b ip ▯▯▯ ▯▯▯ ▯▯▯ a2i + b i1 ▯ a2i + b i2 … ▯ a2i + b ip ▯▯▯ ▯▯▯ ▯▯▯ ▯▯▯ ani + b i1 ani + b i2 … ▯▯▯ ani + b ip ▯▯▯ *all rows must correspond to the same number of columns in the second matrix § Theorem: o If A = 1 1 1 and c = c 1 1 2 an 2 1 1 1 n c then Ac = c a1 1 c a2 2…+ c a n n § A square matrix 2 o AA = A o A A = A 3 § Identity matrix o AI = A o IA = A o AI = IA § More properties 1. A(BC) = (AB)C 2. A(B + C) = AB + AC 3. (A + B)C = AC + AB 4. AI = IA = A 5. (A ) = A T T T 6. (A + B) = B A
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