Solutions to Sample Problems for Test 3

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Marjory Adams
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1 22 Differential Equation Intructor: Petronela Radu November 8 25 Solution to Sample Problem for Tet 3 For each of the linear ytem below find an interval in which the general olution i defined (a) x = x + 2 co t y (b) (t + )u = y = (ln t)x 3 ty; t 3 t u 3v v = (in t)u (co t)v Solution We need continuity of the coefficient in the interior of an interval which contain the initial condition Since here we do not have initial condition we pecify where the data can be given o we are guaranteed exitence and uniquene (a) (dicued in cla) The olution exit on ( π/2) or on (π/2 3) depending on where the initial condition are given (b) The olution exit on ( ) ( 3) or (3 ) depending on where the initial time i taken 2 For the differential ytem x (t)=ax(t): (a) Perform a phae plane analyi; (b) Find the general olution; (c) Dicu the tability of the origin baed on part (a) and (b); (d) Draw ome trajectorie to illutrate what type of a critical point the origin i where 3 2 (a) A = (b) A = (c) A = (d) A = Solution: (a) c) The characteritic equation i (3 λ)(8 λ) 6 = with root 2 and 9 The eigenvector 2 v correponding to λ = 2 atifie v + 2v 2 = o we take v = The eigenvector w correponding to λ = 9 atifie 3w w 2 = o one can take w = The general olution 3 i given by: x(t) = c e 2t 2 + c 2 e 9t 3 o the origin i a ource (hence untable critical point) and an improper node (b) c) The characteritic equation i ( λ)(2 λ) 3 = with root -4 and 7 The eigenvector 6 v correponding to λ = 4 atifie v + 2v 2 = o we take v = The eigenvector w 5

here we do not have initial condition we pecify where the data can be given o we are guaranteed exitence and uniquene (a) (dicued in cla) The olution exit on ( π/2) or on (π/2 3) depending on where

2 correponding to λ = 7 atifie 6w +6w 2 = o one can take w = i given by: x(t) = c e 4t o the origin i a addle (hence untable critical point) c 2 e 7t The general olution (c) c) The characteritic equation i ( λ)(3 λ) + = with complex conjugate root 2 + 3i and 2 3i The eigenvector v correponding to λ = 2 + 3i atifie ( 3i)v + 5v 2 = o we take 5 v = We compute the real and the imaginary part of e + 3i (2+3i)t v and obtain e 2t (co t + i in 3t) v = e 2t ( The general olution i given by: x(t) = c e 2t 5 co 3t co 3t 3 in 3t 5 co 3t co 3t 3 in 3t + c 2 e 2t + o the origin i a piral ource (hence untable critical point) 5 in 3t in 3t + 3 co 3t 5 in 3t in 3t + 3 co 3t (d) c) The characteritic equation i (3 λ)(2 λ) 2 = with root 3 and 2 with multiplicity 2 The eigenvector v correponding to λ = 3 atifie v + 5v 2 + 7v 3 = and v 2 + 9v 3 = By 52 olving the ytem we get v = 52v 3 v 2 = 9v 3 o one can take v = 9 The eigenvector w correponding to λ = 2 atifie 5v 2 + 7v 3 = v 3 = o take w = Since we have only two eigenvector we need to take the generalized eigenvector ū correponding to λ = 2 atifie 5u 2 = o one can take w = /5 The general olution i given by: x(t) = c e 3t c 2 e 2t (t The origin i a ource (hence untable critical point) + /5 3 Decide if the following tatement are TRUE or FALSE Motivate your anwer (a) A center i a table point (b) A center i an aymptotically table point (c) A node can be a ink a ource or a addle (d) A addle point i an aymptotically untable point but ome trajectorie move toward it For the next tatement aume that the origin i the only critical point of x = Ax (e) If A ha real negative eigenvalue then the origin i a ink (f) If the origin i a ource then all trajectorie that tart outide the origin are unbounded (g) If one of the eigenvalue ha the real part equal to then the origin i a center Solution: - dicued in cla 4 Solve the following initial value problem: ) )

e + 3i (2+3i)t v and obtain e 2t (co t + i in 3t) v = e 2t ( The general olution i given by: x(t) = c e 2t 5 co 3t co 3t 3 in 3t 5 co 3t co 3t 3 in 3t + c 2 e 2t + o the origin i a piral ource (hence

3 (a) x = 5x y y = 3x + y with x() = 2 y() = The eigenvalue and correponding eigenvector are: λ = 2 v = 3 The general olution i c = 3/4 c 2 = 5/4 x(t) λ 2 = 4 v 2 = = c e 2t + c 3 2 e 4t (b) x = x 5y y = x 3y with x() = y() = The eigenvalue are ± 2i For + 2i we get the eigenvector 5 2 i The general olution i: x(t) = e t (c 5 co t 2 co t in t From the initial condition c = /5 c 2 = 3/5 + c 2 From initial condition we get 5 in t 2 in t co t (c) x = 3x + 9y y = x 3y with x() = 2 y() = 4 The eigenvalue i ha multiplicity 2 and the correponding eigenvector 3 ) A generalized eigenvector i The generalized olution i x(t) 3 = c 3 + c 2 (t + ) From the initial condition we get c = /2 c 2 = 7/2 5 The following model can be interpreted a decribing the interaction of two pecie with population denitie x and y: x = x 5y y = 25x + y Ue a phae plane analyi to determine the long-time behavior of a olution whoe trajectory pae through the point () Solution: dicued in cla 6 Detrmine if the ytem below could be interpreted a model for predatory-prey competing or cooperating pecie Motivate your anwer (a) x = x( x + y) (b) x = x( x y) (c) x = x( x + y) y = y(4 3y x) y = y(4 3y x) y = y(4 3y + x) Solution: We look for the ign of the other term in an equation (a) Predator-prey x =predator y= prey Both x y have logitic growth (b) Competitive (c) Cooperating

initial condition we get 5 in t 2 in t co t (c) x = 3x + 9y y = x 3y with x() = 2 y() = 4 The eigenvalue i ha multiplicity 2 and the correponding eigenvector 3 ) A generalized eigenvector i The

4 7 Conider two interconnected tank uch that there i a tranfer of mixture between the tank in both direction through two pipe Tank initially contain 3 gal of water and 25 oz of alt and Tank 2 initially contain 2 gal of water and 5 oz of alt Water containing oz/gal of alt flow into Tank at a rate of of 5gal/min The mixture flow from Tank to Tank 2 at a rate of 3 gal/min Water containing 3 oz/gal of alt alo flow into Tank 2 at a rate of gal/min (from the outide) The mixture drain from Tank 2 at a rate of 4 gal/min of which ome flow back into Tank at a rate of 5 gal/min while the remainder leave the ytem (a) Let Q (t) and Q 2 (t) repectively be amount of alt in each tank at time t Write down differential equation and initial condition that model the flow proce Oberve that the ytem in nonhomogeneou (b) Find the value of Q and Q 2 for which the ytem i in equilibrium and denote them by Q E and Q E 2 Can you predict which tank will approach it equilibrium tate more rapidly? (c) Let x (t) = Q (t) Q E and x 2 (t) = Q 2 (t) Q E 2 Determine an initial value problem for x and x 2 Oberve that the ytem i homogeneou Solution: (a) (b) Critical point are given by: Q = 5 3 Q 3 + 5Q 2 2 Q = 25 Q 2 = Q 3 4Q 2 2 Q 2 = Q 3 + 5Q 2 2 = Q 3 4Q 2 2 = Therefore the equilibrium olution are Q E = 42 Q E 2 = 36 Compute Q and Q 2 at the initial value o Q () = 625 Q 2() = 25 The econd tank will approach equilibrium more rapidly (c) With the change of variable the ytem become: x (t) = x + 3x 2 4 x () = 7 x 2 = x x 2 5 x 2() = 2 8 Compute the Laplace tranform of the function { in t t < π f(t) = 3 t t π in two way: by uing the definition and by uing the Heaviide (unit tep) function Solution By the definition we have that Lf() = e t f(t)dt = π e t in tdt + For both integral we ue integration by part and obtain: Lf() = 2 + (e π + ) + e π ( 3 π 2 ) In order to compute the Laplace tranform uing table we write: f(t) = in t( H(t π)) + (3 t)h(t π) π e t (3 t)dt

alt alo flow into Tank 2 at a rate of gal/min (from the outide) The mixture drain from Tank 2 at a rate of 4 gal/min of which ome flow back into Tank at a rate of 5 gal/min while the remainder leave

5 So Lf() = 2 Lin th(t π)() + 3LH(t π)() LtH(t π)() = e π Lin(t + π)() + 3 e π e π Lt + π() Since in(t + π) = in t we get Lf() = e π e π e π (/ 2 + π/) 9 Ue the table to find the invere Laplace tranform of Solution e L = 4 L 2 + 9/ L 2 + 9/4 = co 3 2 t in 3 2 t From the table we have that L e 3 = /4H( 3) 4 where H i the Heaviide (unit-tep) function

tranform of Solution + 2 4 2 + 9 e 3 4 2 + 2 + 5 L + 2 4 2 + 9 = 4 L 2 + 9/4 + 2 4 L 2 + 9/4 = 2 4 3 co 3

Math 22B, Homework #8 1. y 5y + 6y = 2e t

Math 22B, Homework #8 1. y 5y + 6y = 2e t Math 22B, Homework #8 3.7 Problem # We find a particular olution of the ODE y 5y + 6y 2e t uing the method of variation of parameter and then verify the olution uing the method of undetermined coefficient.