Solutions to Assignment 1


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1 MTHE 237 Fll 2015 Solutions to Assignment 1 Problem 1 Find the order of the differentil eqution: t d3 y dt 3 +t2 y = os(t. Is the differentil eqution liner? Is the eqution homogeneous? b Repet the bove for the following differentil eqution: d 2 y dt 2 +3y dt = 0. Consider the eqution y 3y +2y = 0, By trying solution of the form y = e rt, find the possible r vlues stisfying the differentil eqution. The order is 3, nd the eqution is liner. The eqution is nonhomogeneous. b The eqution is nonliner (The reson here is tht yy is not liner funtion of (y,y. The eqution hs order 2. The eqution is homogeneous. By substituting y = e rt, nd noting tht y = re rt nd y = r 2 e rt, it follows tht e rt (r 2 3r+2 = 0, sine e rt 0, it follows tht (r 2(r 1 = 0, whih implies tht r = 1 or r = 2. Problem 2 Consider the following differentil eqution where > 0,b > 0 re positive relvlued numbers. dt = y+b, Obtin the generl solution of the differentil eqution. b Sketh the prtiulr solutions for three different initil onditions, tht is for three different y(0 vlues. Desribe how the solutions hnge if (i inreses (ii b inreses. (iii b = 0 nd inreses. d Drw qulittive diretion field for the eqution dt = 5y+10 for t 0. In your grph, highlight the prtiulr solution pssing through the point (t,y = (0,10. The eqution dt = y +b
2 n be written s y +b = dt It follows by multiplying boths sides by nd integrtion tht for some onstnt C. Hene, if y > b/, we hve nd if y < b/, we hve with K = e C. b,,dif y(0 = y 0, then K = b/ y 0. Thus, ln( y b/ = t+c, y(t = b/+ke t, y(t = b/ Ke t, y(t = b/+(y 0 b/e t, If inreses, the equilibrium vlue is pprohed more rpidly, the equilibrium vlue is b/. Problem 3 Aording to Newton s lw of ooling, the temperture t time t, u(t, of n objet stisfies the following differentil eqution: du dt = k(u T, where T is the onstnt temperture of the environment nd k > 0 is positive onstnt. Let the initil ondition be suh tht u(0 = u 0. Find the temperture of the objet s funtion of t. b Let τ > 0 be the time t whih the initil temperture differene between u(t nd T, u 0 T hs been redued by hlf, tht is u(τ T = u 0 T. 2 Find the reltion between k nd τ. We hve with u(0 = u 0. It follows tht Hene, du dt = k(u T ln( u T = kt+c. u = Ke kt +T for some onstnt K. Sine u(0 = u 0, we hve tht K = u 0 T. Hene, u(t = (u 0 Te kt +T b We hve tht hene, e kτ = 1/2 nd u(τ T = 1 2 (u 0 T, kτ = ln(2 Problem 4 Consider the following differentil eqution: (2x+4xydx+(2x 2 +3y 2 = 0,
3 defined on x 0,y > 0 Is the eqution ext? b Obtin generl (impliit solution to the eqution: Find the prtiulr impliit solution whih psses through the point (x,y = (0,5. It follows tht Hene, the eqution is ext. M(x,y = 2x+4xy N(x,y = 2x 2 +3y 2 y M(x,y = N(x,y = 4x x b Due to extness, by theorem tht we proved in lss, there exists funtion F(x,y suh tht F x = M(x,y nd F y = N(x,y Upon integrting M(x,y long the x oordinte we obtin: F(x,y = x 2 +2x 2 y +g(y+c for some g(. whih is funtion of y only. Now, tking the prtil derivtive of this term with respet to y nd equting it to N, it follows tht g (y = 3y 2, whih leds to g(y = y 3 plus some onstnt term, whih is lre ounted for by C. Hene, F(x,y = x 2 +2x 2 y +y 3 +C = 0, is generl solution. To find the prtiulr solution t (1,1, we substitute the vlues nd obtin tht C = 4. As suh, the prtiulr solution is x 2 +2x 2 y +y 3 = 4. There is unique prtiulr solution by the existene nd uniqueness theorem sine hs ontinuous prtil derivtives in y for y > 0. /dx = 3x2 +4xy 2x 2 +3y 2, Problem 5 Solve the differentil eqution xdx+y = 0 with initil ondition y(1 = 1. This is n ext differentil eqution sine with M(x,y = x,n(x,y = y, y M(x,y = 0 = x N(x,y. Find F(x,y = x M(s,yds + K(y = x 2 + K(y. Tking the prtil derivtive with respet to y nd equling to N leds to K(y = y 2 /2+. As result F(x,y = (1/2(x 2 +y 2 + = 0 is solution. y(1 = 1 implies tht = 2 (Note tht this is the only solution by the existene nd uniqueness theorem sine when x > 0, y > 0, x/y is ontinuous with ontinuous prtil derivtives. Problem 6 Consider differentil eqution dx = f(x,y, where f(x,y is ontinuous funtion nd f(x,y hs ontinuous prtil derivtives in y, on D R 2 whih is losed nd bounded set. Consider two different solutions orresponding to two different initil onditions (x 0,y 0 nd (x 0,ỹ 0 where (x 0,y 0 D nd (x 0,ỹ 0 D. Suppose, furthermore tht the differentil eqution is defined only on D. Cn these two solutions interset t ny point in D? We will use ontrdition to rrive t the orret onlusion. Suppose tht there exists point ( x,ỹ D suh tht the two solutions pss through this point nd they strt to behve differently fter this point. By
4 the existene nd uniqueness theorem, this implies tht there is unique solution in smll neighborhood of this point. Then, there nnot exist suh point. The solutions re then either identil everywhere; or do not interset t ll. Problem 7 Show tht given funtion F, if x ( yf(x,y nd y ( xf(x,y re ontinuous, they must be equl. This result is known s Clirut s Theorem. Let,b,,d R, < b nd < d. Let F x = x F, F y = x ( y F = F yx nd y ( x F = F yx. Now, by the fundmentl theorem of lulus y F, b d b F xy dx = b ( d = F x (x,d F x (x, dx = F(b,d F(b, F xy dx ( F(,d F(, A prllel rgument pplies for the integrtion d b F xydx leding to d b F yx dx = ( F(b,d F(,d F(b, F(, Now, these two imply tht, by ontinuity, they hve to be equl inside the integrtion s well. Suppose not. Then, for some ǫ > 0, there exists point (x 0,y 0 where F yx (x 0,y 0 F xy (x 0,y 0 ǫ. But then, sine the funtions re ontinuous, there exists neighborhood of this point (sy (x 0 δ,x 0 +δ (y 0 δ,y 0 +δ for some δ > 0 where the funtions re different by t lest ǫ/2. Then, the integrl of the funtions F xy dx nd F yx dx on this neighborhood will be stritly different. Hene, ontrdition. It must be tht F xy = F yx, if they re ontinuous. (1 Some dditionl problems: Problem 8 Consider the fmily of prbols defined by y = kx 2, where k is n rbitrry onstnt. Find the fmily of urves whih intersets the given fmily of prbols orthogonlly t eh point. You re sked to find n expression in the form F(x,y = 0 (this my inlude onstnt term suh tht the fmily of suh urves is orthogonl to the fmily of prbols for every k. Hint: Use the ft tht line with slope m t point P, is orthogonl to the line with slope 1/m pssing through the sme point. We hve tht the solution should stisfy dx = 1 2kx = 1 2(y/x 2 x = 1 2 x/y,
5 Hene, xdx+2y = 0 This eqution is ext nd the solution is given by the ellipsoid: x 2 +2y 2 =, for n rbitrry onstnt. Problem 9 Obtin the generl solution to the differentil eqution for x > 0,y > 0. (3x 2 +4xydx+(2x 2 +3y 2 = 0, This is n ext differentil eqution sine with M(x,y = 3x 2 + 4xy,N(x,y = 2x 2 + 3y 2, y M(x,y = 4x = x N(x,y. Find F(x,y = x M(s,yds +K(y = x 3 +2x 2 y + K(y. Tking the prtil derivtive with respet to y nd equling to N leds to K(y = y 3 +. As result F(x,y = x 3 +2x 2 y+y 3 = K is the generl solution.
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