Elementary Differential Equations and Boundary …

1 September 11, 2008 11:18 boyce-9e-bvp Sheet number 108 Page number 88cyanblack88 Chapter 2. First Order Differential Equationsplace to permit successful breeding, and the population rapidly declined to extinc-tion. The last survivor died in 1914. The precipitous decline in the passenger pigeonpopulation from huge numbers to extinction in a few decades was one of the earlyfactors contributing to a concern for conservation in this 1 through 6 involve Equations of the formdy/dt=f(y). In each problem sketchthe graph off(y)versusy, determine the critical (equilibrium) points, and classify each one asasymptotically stable or unstable. Draw the phase line, and sketch several graphs of solutionsin +by2,a>0,b>0,y0 +by2,a>0,b>0, <y0< (y 1)(y 2),y0 1, <y0< y 1, <y0< 2(arctany)/(1+y2), <y0< Equilibrium a constant equilibrium solution has theproperty that solutions lying on one side of the equilibrium solution tend to approach it,whereas solutions lying on the other side depart from it (see Figure ).

2 In this case theequilibrium solution is said to (b)kyt(a)(t) = k (t) = k FIGURE both cases the equilibrium solution (t)=kis semistable.(a)dy/dt 0; (b)dy/dt 0.(a) Consider the equationdy/dt=k(1 y)2,(i)wherekis a positive constant. Show thaty=1 is the only critical point, with the corre-sponding equilibrium solution (t)=1.(b) Sketchf(y)versusy. Show thatyis increasing as a function oftfory<1 and alsofory>1. The phase line has upward-pointing arrows both below and abovey=1. Thussolutions below the equilibrium solution approach it,and those above it grow farther (t)=1 is semistable.(c) Solve Eq. (i) subject to the initial conditiony(0)=y0and confirm the conclusionsreached in part (b).Problems 8 through 13 involve Equations of the formdy/dt=f(y).

3 In each problem sketchthe graph off(y)versusy, determine the critical (equilibrium) points, and classify each oneSeptember 11, 2008 11:18 boyce-9e-bvp Sheet number 109 Page number Autonomous Equations and Population Dynamics89as asymptotically stable, unstable, or semistable (see Problem 7). Draw the phase line, andsketch several graphs of solutions in k(y 1)2,k>0, <y0< (y2 1), <y0< (1 y2), <y0< b y,a>0,b>0,y0 (4 y2), <y0< (1 y)2, <y0< 14. Consider the equationdy/dt=f(y)and suppose thaty1is a critical point,that is,f(y1)= that the constant equilibrium solution (t)=y1is asymptotically stable iff (y1)<0and unstable iff (y1)> Suppose that a certain population obeys the logistic equationdy/dt=ry[1 (y/K)].(a) Ify0=K/3,find the time at which the initial population has doubled.

4 Find the valueof corresponding tor= per year.(b) Ify0/K= , find the timeTat whichy(T)/K= , where 0< , <1. Observe thatT as 0oras 1. Find the value ofTforr= per year, = , and = Another equation that has been used to model population growth is the Gompertz14equationdy/dt=ryln(K/y),wherer andKare positive constants.(a) Sketch the graph off(y)versusy, find the critical points, and determine whether eachis asymptotically stable or unstable.(b) For 0 y K, determine where the graph ofyversustis concave up and where it isconcave down.(c) For eachyin 0<y K, show thatdy/dtas given by the Gompertz equation is neverless thandy/dtas given by the logistic (a) Solve the Gompertz equationdy/dt=ryln(K/y),subject to the initial conditiony(0)= :You may wish to letu=ln(y/K).

5 (b) For the data given in Example 1 in the text (r= per year,K= 106kg,y0/K= ), use the Gompertz model to find the predicted value ofy(2).(c) For the same data as in part (b), use the Gompertz model to find the time at whichy( )= A pond forms as water collects in a conical depression of radiusaand depthh. Supposethat water flows in at a constant ratekand is lost through evaporation at a rate proportionalto the surface area.(a) Show that the volumeV(t)of water in the pond at timetsatisfies the differentialequationdV/dt=k (3a/ h)2/3V2/3,where is the coefficient of Gompertz (1779 1865) was an English actuary. He developed his model for populationgrowth, published in 1825, in the course of constructing mortality tables for his insurance 11, 2008 11:18 boyce-9e-bvp Sheet number 119 Page number Exact Equations and Integrating Factors99 Thus, if( M)yis to equal( N)x, it is necessary thatd dx=My NxN.

6 (27)If(My Nx)/Nis a function ofxonly, then there is an integrating factor that alsodepends only onx; further, (x)can be found by solving Eq. (27),which is both linearand similar procedure can be used to determine a condition under which Eq. (23)has an integrating factor depending only ony; see Problem an integrating factor for the equation(3xy+y2)+(x2+xy)y =0(19)and then solve the Example 3 we showed that this equation is not exact. Let us determine whether it has anintegrating factor that depends onxonly. On computing the quantity(My Nx)/N, we findthatMy(x,y) Nx(x,y)N(x,y)=3x+2y (2x+y)x2+xy=1x.(28)Thus there is an integrating factor that is a function ofxonly, and it satisfies the differentialequationd dx= x.(29)Hence (x)=x.(30)Multiplying Eq.

7 (19) by this integrating factor, we obtain(3x2y+xy2)+(x3+x2y)y =0.(31)The latter equation is exact, and its solutions are given implicitly byx3y+12x2y2=c.(32)Solutions may also be found in explicit form since Eq. (32) is quadratic may also verify that a second integrating factor for Eq. (19) is (x,y)=1xy(2x+y),and that the same solution is obtained, though with much greater difficulty, if this integratingfactor is used (see Problem 32).PROBLEMSD etermine whether each of the Equations in Problems 1 through 12 is exact. If it is exact, findthe (2x+3)+(2y 2)y =02.(2x+4y)+(2x 2y)y =03.(3x2 2xy+2)dx+(6y2 x2+3)dy=04.(2xy2+2y)+(2x2y+2x)y =0 September 11, 2008 11:18 boyce-9e-bvp Sheet number 120 Page number 100cyanblack100 Chapter 2.

8 First Order Differential ax+bybx+ ax bybx cy7.(exsiny 2ysinx)dx+(excosy+2 cosx)dy=08.(exsiny+3y)dx (3x exsiny)dy=09.(yexycos 2x 2exysin 2x+2x)dx+(xexycos 2x 3)dy=010.(y/x+6x)dx+(lnx 2)dy=0,x>011.(xlny+xy)dx+(ylnx+xy)dy=0;x >0,y> (x2+y2)3/2+ydy(x2+y2)3/2=0In each of Problems 13 and 14 solve the given initial value problem and determine at leastapproximately where the solution is (2x y)dx+(2y x)dy=0,y(1)=314.(9x2+y 1)dx (4y x)dy=0,y(1)=0In each of Problems 15 and 16 find the value ofbfor which the given equation is exact, andthen solve it using that value (xy2+bx2y)dx+(x+y)x2dy=016.(ye2xy+x)dx+b xe2xydy=017. Assume that Eq. (6) meets the requirements of Theorem in a rectangleRand istherefore exact. Show that a possible function (x,y)is (x,y)= xx0M(s,y0)ds+ yy0N(x,t)dt,where(x0,y0)is a point Show that any separable equationM(x)+N(y)y =0is also each of Problems 19 through 22 show that the given equation is not exact but becomes exactwhen multiplied by the given integrating factor.

9 Then solve the +x(1+y2)y =0, (x,y)=1/xy320.(sinyy 2e xsinx)dx+(cosy+2e xcosxy)dy=0, (x,y)= +(2x yey)dy=0, (x,y)=y22.(x+2)sinydx+xcosydy=0, (x,y)=xex23. Show that if(Nx My)/M=Q, whereQis a function ofyonly, then the differentialequationM+Ny =0has an integrating factor of the form (y)=exp Q(y) Show that if(Nx My)/(xM yN)=R, whereRdepends on the quantityxyonly, thenthe Differential equationM+Ny =0has an integrating factor of the form (xy). Find a general formula for this 11, 2008 11:18 boyce-9e-bvp Sheet number 164 Page number 144cyanblack144 Chapter 3. Second Order Linear EquationsPROBLEMSIn each of Problems 1 through 8 find the general solution of the given Differential +2y 3y= +3y +2y=03. 6y y y= 3y +y= +5y = 9y= 9y +9y= 2y 2y=0In each of Problems 9 through 16 find the solution of the given initial value problem.

10 Sketchthe graph of the solution and describe its behavior +y 2y=0,y(0)=1,y (0)= +4y +3y=0,y(0)=2,y (0)= 111. 6y 5y +y=0,y(0)=4,y (0)= +3y =0,y(0)= 2,y (0)= +5y +3y=0,y(0)=1,y (0)=014. 2y +y 4y=0,y(0)=0,y (0)= +8y 9y=0,y(1)=1,y (1)=016. 4y y=0,y( 2)=1,y ( 2)= 117. Find a Differential equation whose general solution isy=c1e2t+c2e Find a Differential equation whose general solution isy=c1e t/2+c2e the solution of the initial value problemy y=0,y(0)=54,y (0)= the solution for 0 t 2 and determine its minimum Find the solution of the initial value problem2y 3y +y=0,y(0)=2,y (0)= determine the maximum value of the solution and also find the point where thesolution is Solve the initial value problemy y 2y=0,y(0)= ,y (0)=2.