DIFFERENTIAL EQUATIONS PRACTICE PROBLEMS: ANSWERS

1 DIFFERENTIAL EQUATIONS PRACTICE PROBLEMS: ANSWERS1. Find the solution ofy0+2xy=x,withy(0) = is a linear equation. The integrating factor iseR2xdx=ex2. Multiplying through by this, we gety0ex2+2xex2y=xex2(ex2y)0=xex2ex2y=Rxe x2dx=12ex2+Cy=12+Ce in the initial condition givesC= 5/2,soy=12 52e= Find the general solution ofxy0=y (y2/x).A number of substitutions will work here. The simplest isy=ux,soy0=u+ gives a separable equation:x(u+u0x)=ux u2x2x=ux u2xdudxx2= u2xZ u 2du=Z1xdx1u=lnx+Cu=1lnx+Cy=xlnx+ Suppose that the frog populationP(t)of a small lake satisfies the DIFFERENTIAL equationdPdt=kP(200 P).(a) Find the equilibrium solutions. Sketch them and using the equation, sketch several solution curves,choosing some with initial points above and between the equilibrium equilibrium solutions areP=0(unstable) andP= 200(stable).

2 1(b) In the year 2000, its population was 100 and growing at the rate of 5 per year. Predict the lake s frogpopulation in 2008. Note:1P(200 P)=1/200P+1/200(200 P). This is a separable equation:Z1P(200 P)dP=ZkdtZ1200 1P+1200 P =Zkdt1200(ln(P) ln(200 P)) =Zkdtln(200 P) ln(P)=ln 200 PP = 200kt+C200P 1=Ke the base year, the initial conditionP(0) = 100givesK= 100anddPdt t=0=5givesk=510, :200P 1=e 200510,000(8)=e 4/5P=2001+e 4/5 Find the general solution of the DIFFERENTIAL equationy00 y0=ex the DIFFERENTIAL operatorD, the homogeneous equationy00 y0=0becomesD2 D=0which hassolutionsD=1andD=0, corresponding toDy=y(y=ex)andDy=0(y=constant). Thus, thegeneral solution to the homogeneous equation isyh=c1+ a particular solution to theoriginal equation using undetermined coefficients.

3 Our guess might beyp=Aex+Bx2+Cx+D,Butexduplicates part of the homogeneous solution as does the derivative ofCx(the constantc1). So we multiplyby a high enough power ofxto avoid this;xwill do:yp=Axex+Bx3+Cx2+Dxy0p=Axex+Aex+3Bx2+2 Cx+Dy00p=Axex+2 Aex+6Bx+2Cy00p y0p=Aex 3Bx2+(6B 2C)x+(2C D).We set this equal toex 9x2,whichgives:A=1,B=3,C=9andD=18. Since the general solution toa linear DE is the general solution to the associated homogeneous equation + a particular solution to theoriginal, the general solution isy=c1+c2ex+xex+3x3+9x2+ A mass of2kg is attached to a spring with constantk=8 Newtons/meter.(a) Find the natural frequency of this system equation (no driving force) is2x00+8x=0orx00+4x= +4=0soD= ,thesolutionisx(t)=c1cos 2t+c2sin 2t, and the frequency is 0=2(radians persecond or1/ hertz).

4 (b) If the motion is also subject to a damping force withc=4 Newtons/(meter/sec), and the mass isinitially pulled 1 meter beyond its equilibrium point and released (without initial velocity),find themotion,x(t).(Youmayleaveyouranswerina nyform.)2We could use Laplace methods here, but we ll use theDoperator again. The equation2x00+4x0+8x=0, which becomesD2+2D+4=0,havingrootsD= 2 4 162= 1 :x(t)=e t c1cos 3t +c2sin 3t x0(t)= e t c1cos 3t +c2sin 3t +e t c1 3sin 3t +c2 3cos 3t .The initial conditionsx(0) = 1,x0(0) = 0now givec1=1,c2=1/ 3,sox(t)=e t cos 3t +(1/ 3) sin 3t 6. Findand sketchthe solution to the initial value problemy00+4y= (t ),y(0) =y0(0) = the Laplace transform givess2Y+4Y=e s,soY(s)=e s 1s2+4.

5 Now1s2+4=12 2s2+4 untransforms into12sin 2t,soY(s)untransforms into:y(t)=u(t )12sin Find the inverse Laplace transform ofF(s)=s2+4s(s2+1)=As+Bs+Cs2+1,sos2+4=A s2+1 +(Bs+C) 1gives the equationsB+C= 3andB C= 3,soB= 3andC= ,F(s)=4s 3ss2+1f(t)=4 LetA= 143155251 (a) FindA 1, put the identity next toAand row reduce the augmented matrix: 143155251 143155251 100010001 A 1z}|{ 20 11 59 5 2 531 3(b) Use your answer above to solveA x= bwhere b=(1,0, 1). x1x2x3 = 20 11 59 5 2 531 10 1 = 2511 6 .9. The matrixB= 022202220 has eigenvalues =4, 2. Find a basis of row reduceB Ifor =4and = 2:B 4I= 42 22 4222 4 10 101 100 0 ,z=sy=z=sx=z=sB+2I= 222222222 111000000 ,z=sy=tx= s tFor =4,lettings=1gives the eigenvector(1,1,1)(as a column); for = 2,lettings=0,t= 1gives(1, 1,0)and lettings= 1,t=0gives(1,0, 1).

6 Eigenvectors for distinct eigenvalues are alwaysindependent, and the two vectors for the eigenvalue = 2are clearly independent (neither is a multipleof the other). Thus, these three vectors are a basis for the eigenspace ofB; The reduced row echelon form for the matrixAbelow has been computed by Matlab:A= 2 4 12 36 1 55 10 4 1 rref(A)= 1 20300140000 Use this tofind all solutions of2x1 4x2 x3=2 3x1+6x2+x3= 55x1 10x2 4x3= 1and express your answer in vector of the row-reduced matrix as an augmented matrix we see that there is no restriction onx2,soletx2=s. The second row saysx3=4and thefirst row saysx1 2x2=3orx1=3+ have: x1x2x3 =s 210 + 304 .11. Letv1=(2,1,3),v2=(1,5,9),andw=(1, 1, 1).

7 Iswin span(v1,v2)? Find a basis for span(v1,v2,w).What is thedimensionof span(v1,v2,w)?We make these vectors into the column of a matrixA. A linear dependence among the vectors is then asolution to the equationAX= (X6=0)one: 21 115 139 1 10 2/301 1/300 0 ,z=sy=(1/3)sx= (2/3)sThus, (2/3)v1+(1/3)v2+w=0is a non-trivial dependency, allow us to solve forwin terms ofv1andv2. So Span(v1,v2,w)=Span(v1,v2).Sincev1andv2ar e clearly not multiples of one another, they areindependent, hence form a basis for Span(v1,v2).Thus,Span(v1,v2,w)has Consider the following system offirst-order DIFFERENTIAL EQUATIONS :x01=9x1+5x2x1(0) = 1x02= 6x1 2x2x2(0) = 0 Use eigenvalues and eigenvectors tofind the matrix form these EQUATIONS become x01x02 =Az}|{ 95 6 2 x1x2 The characteristic polynomial forAisdet(A xI)= 9 x5 6 2 x =x2 7x+12=(x 3)(x 4),sothe eigenvalues are =3, 3I= 65 6 5 15/600 which givesv3= 5 6 A 4I= 55 6 6 1100 which givesv4= 1 1 So the general solution is x1x2 =c1 56 e3t+c2 1 1 , the initial conditions give 10 = 5c1 6c1 + c2 c2 or the equations6c1+c2=0and5c1+c2=1, with solutionsc1= 1andc2= gives the solutionx1= 5e3t+6e4tx2=6e3t Here is a sawtooth functionf(t).

8 T1234561f(t)f(t) =tThefirst tooth is the functionf1(t)= tfor0 t< (a) From the definitionL{f}(s)=F(s)=Z 0f(t)e stdt, show thatF1(s)=1 e ss2 e ss.(Useintegration by parts; you only have to integrate from0to1.)5F1(s)=Z 0f1(t)e stdt=Z10te stdt. We use the partsu=t,anddv=e stdt:Zudv=uv Zvdu= tse st+1sZe stdt= tse st 1s2e stdt= tse st 1s2e st 10= 1se s 1s2e s 0 1s2 =1s2 1 e s 1se s.(b) For a periodic functionfof periodp,F(s)=11 e psZp0f(t)e stdt. Use this and part (a) to showthat, for the sawtooth:F(s)=1s2 e ss(1 e s).This is simple