20D - Homework Assignment 0 - UCSD Mathematics

1 Brian Bowers (TA for Hui Sun)MATH 20 DHomework Assignment 0 October 1, 201320D - Homework Assignment #1-6 Determine the order of the given differential equation; also state whether the equation is linearor nonlinear.(1)t2d2ydt2+tdydt+ 2y= sint(2) (1 +y2)d2ydt2+tdydt+y=et(3)d4ydt4+d3ydt3+d2 ydt2+dydt+y= 1(4)dydt+ty2= 0(5)d2ydt2+ sin(t+y) = sint(6)d3ydt3+tdydt+ (cos2t)y=t3(1) The highest derivative is a second derivative, so the order is 2. The equation matches the form ofequation (11) on page 21, so it is linear.(2) The highest derivative is a second derivative, so the order is 2. The term (1 +y2)d2ydt2 means that thisis a nonlinear equation since (1 +y2) is not a function oft.

2 (3) The highest derivative is a fourth derivative, so the order is 4. The equation matches the form of equation(11) on page 21, so it is linear.(4) The highest derivative is a first derivative, so the order is 1. The term ty2 means that this is a nonlinearequation sincetyis not a function oft.(5) The highest derivative is a second derivative, so the order is 2. The term sin(t+y) means that this isa nonlinear equation since sin(t+y) is not a function oft.(6) The highest derivative is a second derivative, so the order is 3. The equation matches the form ofequation (11) on page 21, so it is # 7,11 Verify that each given function is a solution of the differential equation.

3 (7)y y= 0;y1(t) =et,y2(t) = cosht(11) 2t2y + 3ty y= 0,t >0;y1(y) =t1/2,y2(t) =t 1To verify that a function is a solution to a differential equation, all we need to do is plug it into thedifferential equation and check that the equation is true.(7) First, we plug iny1fory:y 1 y1= (et) (et)= (et) (et)=et et= 0 Next, we plug iny2fory. This function involves cosh, a hyperbolic trig function. All we need to knowfor the purposes of this course is that (sinh) = cosh and (cosh) = 2 y2= (cosht) cosht= (sinht) cosht= cosht cosht= 0 Thus,y1andy2are solutions to the differential equation.(11) First, we plug iny1fory:2t2y + 3ty y= 2t2(t1/2) + 3t(t1/2) (t1/2)= 2t2(12t 1/2) + 3t(12t 1/2) t1/2= 2t2( 14t 3/2)+ 3t(12t 1/2) t1/2= 12t1/2+32t1/2 t1/2= 0 Next, we plug + 3ty y= 2t2(t 1) + 3t(t 1) (t 1)= 2t2( t 2) + 3t( t 2) t 1= 2t2(2t 3) + 3t( t 2) t 1= 4t 1 3t1 t 1= 0 Thus,y1andy2are solutions to the differential #15,17 Determine the values ofrfor which the given differential equation has solutions of the formy=ert.

4 (15)y + 2y= 0(17)y +y 6y= 0(15) We plugy=ertinto the differential equation and solve:y + 2y= (ert) + 2(ert)=rert+ 2ert= (r+ 2)ertBut we know from the original differential equation that this is equal to 0, (r+ 2)ert= set each factor equal to zero and solve.(r+ 2) = 0 = r= 2ert= 0 = no solutionSo, our only solution isr= 2.(17) We plugy=ertinto the differential equation and solve:y +y 6y= (ert) + (ert) 6(ert)= (rert) + (rert) 6(ert)= (r2ert) + (rert) 6(ert)= (r2+r 6)ertBut we know from the original differential equation that this is equal to 0, (r2+r 6)ert= set each factor equal to zero and +r 6 = 0 = (r+ 3)(r 2) = 0 = r= 3,2ert= 0 = no solutionSo, our only solutions arer= 3 andr= #19 Determine the values ofrfor which the given differntial equation has solutions of the formy=trfort > + 4ty + 2y= 0We ll plugy=trinto the equation and simplify.

5 T2y + 4ty + 2y=t2(tr) + 4t(tr) + 2(tr)=t2(rtr 1) + 4t(rtr 1) + 2(tr)=t2(r(r 1)tr 2) + 4t(rtr 1) + 2(tr)=t2r2tr 2 t2rtr 2+ 4trtr 1+ 2tr=r2tr rtr+ 4rtr+ 2tr= (r2+ 3r+ 2)trBut we know from the original differential equation that this is equal to 0, (r2+ 3r+ 2)tr= se each factor equal to zero and + 3r+ 2 = (r+ 2)(r+ 1) = 0 = r= 2, 1tr= 0 = no solution [sincet >0]Thus, the only solutions arer= 2 andr= #1c,3c,4c,6cFor each problem, find the general solution of the given differential equation, and useit to determine how solutions behave ast .(1c)y + 3y=t+e 2t(3c)y +y=te t+ 1(4c)y + (1/t)y= 3 cos 2t(6c)ty + 2y= sintBefore we get into specific problems, I ll give a brief overview of two important and useful of Integrating Factors:First, I will review the Method of Integrating Factors.

6 We only use thismethod when we have an equation of the formy +p(t)y=g(t). The goal is to multiply the entireequation by some unknown function and then figure out what should be to make the left side ofthe equation equal to ( y) . We can recall the product rule to see that( y) = y + our original equation by , we get y +p(t)y = g(t).We see that the y matches up with the equation above, but we still need to take care of the yportion. Thus, we conclude that y=p(t)y = =p(t) [separating variables]= ln = p(t)dt+C[integrating]= ln = p(t)dt[get rid ofC! We want any old integrating factor that works.]= =e p(t)dt[exponentiating]SHORTCUT:So, if you d prefer to skip all the above steps, you can just always use =e p(t)dtandsave yourself some Integration:You CAN complete this course without ever using this tool, but it makes certaintypes of integration MUCH faster, and I hope you ll consider learning and using it.

7 Tabular integrationis essentially a specialized shortcut for integration by parts. We use tabular integration when we wantto integrate an expression with the form(polynomial)(cyclic function).What do I mean by cyclic function? I mean any function that you can integrate some number of timesuntil you end up with a constant times the original our purposes, cyclic functions aresin(at+b),cos(at+b),eat+ are the steps we do every time:(1) Set up a two-column table (hence the name tabular ).(2) We put the polynomial on the top of the left column and the cyclic function on the top of theright (3) For each subsequent row of the table, we differentiate the left side and integrate the right stop making new rows once the left side is equal to 0.

8 (4) Starting with the top-left entry, draw an arrow that moves right one entry and down one entry,repeating until you run out of entries on the right side of the table. Label these arrows alternating +,-,+,-,..(5) Multiply along each arrow (including multiplying by +1 for + and ( 1) for . Add theseproducts together - this is the answer. (Don t forget to add +C .)For example, let s say we wanted to integratet4e2t. With standard integration by parts techniques,this would take a VERY long time. Instead, let s use our new , we should get the integral to bet4(+1)(12e2t)+ 4t3( 1)(14e2t)+ 12t2(+1)(18e2t)+ 24t( 1)(116e2t)+ 24(+1)(132e2t)=12t4e2t t3e2t+32t2e2t 32te2t+34e2tNow let s do the problems.)

9 (1c)The left side of the equation isy + 3y, which corresponds toy +p(t)yin our general form. Thus,p(t) = 3, which means that our integrating factor should be =e p(t)dt=e3t. So, we multiply bothsides of the equation by this (y + 3y) =e3t(t+e 2t)e3ty + 3e3ty=te3t+etRecall that the goal of the integrating factor is to make the left side of the equation into ( y) . Wecan check that this is the case. Thus, we have(e3ty) =te3t+et= e3ty=13te3t 19e3t+et+C[integrating using the tabular integration below]= y=13t 19+e 2t+Ce 3t[dividing bye3t]Below is the work for the tabular integration we did above:6 Moreover, astgets bigger, we see that the13tterm grows, 19, stays constant, and all other terms goto 0.

10 Thus, the solutions will be asymptotic to13t 19ast .(3c)The left side of the equation isy +y, which corresponds toy +p(t)yin our general form. Thus,p(t) = 1, which means that our integrating factor should be =e p(t)dt=et. So, we multiply bothsides of the equation by this (y +y) =et(te t+ 1)ety +ety=t+etRecall that the goal of the integrating factor is to make the left side of the equation into ( y) . Wecan check that this is the case. Thus, we have(ety) =t+et= ety=t22+et+C[integrating]= y=t2e t2+ 1 +Ce t[dividing byet]Moreover, astgets bigger, every term except 1 goes to 0. Thus,yapproaches 1 ast .(4c)The left side of the equation isy + (1/t)y, which corresponds toy +p(t)yin our general form.