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- Question : 1PS - Draw the graph of y = et by hand, for -1 ::; t ::; 1. What is its slope dy / dt at t =0 ? Add the straight line graph of y = et. Where do those two graphs cross ?
- Question : 2PS - Draw the graph of y1 = e2t on top of y2 = 2et. Which function is larger at t =0 ? Which function is larger at t =1 ?
- Question : 3PS - What is the slope of y = e-t at t = 0? Find the slope dy / dt at t = 1.
- Question : 4PS - What "logarithm" do we use for the number t (the exponent) when et = 4?
- Question : 5PS - State the chain rule for the derivative dy / dt if y( t) =f ( u( t)) ( chain off and u).
- Question : 6PS - The second derivative of et is again et. So y = et solves d2y / dt2 = y. A second order differential equation should have another solution, different from y = Get. What is that second solution ?
- Question : 7PS - Show that the nonlinear example dy / dt = y2 is solved by y = C / (1 - Ct) for every constant C. The choice C =1 gave y = 1/(1 - t), starting from y(0) = 1.
- Question : 8PS - Why will the solution to dy / dt = y2 grow faster than the solution to dy / dt = y (if we start them both from y =1 at t = 0)? The first solution blows up at t = 1. The second solution et grows exponentially fast but it never blows up.
- Question : 9PS - Find a solution to dyjdt = -y2 starting from y(0) = 1. Integrate dy/y2 and -dt. (Or work with z = 1/y. Then dz/dt = (dzjdy) (dy/dt) = (-1/y2)(-y2) = 1. From dzjdt =1 you will know z(t) and y = 1/z.)
- Question : 10PS - Which of these differential equations are linear (in y)? (a) y'+ siny=t (b) y' = t2(y - t)
- Question : 11PS - The product rule gives what derivative for ete-t ? This function is constant. At t =0 this constant is 1. Then ete-t =1 for all t.
- Question : 12PS - dy / dt = y + 1 is not solved by y =e t + t. Substitute that y to show it fails. We can't just add the solutions to y' = y and y' = 1. What number c makes y = et + c into a correct solution ?
- Question : 1PS - Set t = 2 in the infinite series for e2. The sum must be e times e, close to 7 .39. How many terms in the series to reach a sum of 7? How many terms to pass 7.3?
- Question : 2PS - Starting from y(O) = 1, find the solution to dy/dt = y at time t = 1. Starting from that y(l), solve dy/dt = -y to time t = 2. Draw a rough graph of y(t) from t =0 tot= 2. What does this say about e-1 times e?
- Question : 3PS - Start with y(O) = $5000. If this grows by dy / dt = .02y until t =5 and then jumps to a = .04 per year until t = 10, what is the account balance at t = 10 ?
- Question : 4PS - Change Problem 3 to start with $5000 growing at dy / dt = .04y for the first five years. Then drop to a = .02 until t = 10. What is now the balance at t = 10 ? Problems 5-8 are about y = eat and its infinite series.
- Question : 5PS - Replace t by at in the exponential series to find eat : at 1 ( )2 1 ( )n e = 1 + at + - at +
- Question : 6PS - Start from y' = ay. Take the derivative of that equation. Take the nth derivative. Construct the Taylor series that matches all these derivatives at t = 0, starting from 1 +at+
- Question : 7PS - At what times t do these events happen? (a) eat= e (b) eat =e2
- Question : 8PS - If you multiply the series for eat in Problem 5 by itself you should get the series for e2at. Multiply the first 3 terms by the same 3 terms to see the first 3 terms in e2at.
- Question : 9PS - (recommended) Find y(t) if dy/dt = ay and y(T) = 1 (instead of y(0) = 1).
- Question : 10PS - (a)If dy / dt = (ln 2)y, explain why y(l) = 2y(0). (b)If dy / dt = -(ln 2)y, how is y(l) related to y(0) ?
- Question : 11PS - In a one-year investment of y(0) = $100, suppose the interest rate jumps from 6% to 10% after six months. Does the equivalent rate for a whole year equal 8%, or more than 8%, or less than 8% ?
- Question : 12PS - If you invest y(0) = $100 at 4% interest compounded continuously, then dy / dt = .04y. Why do you have more than $104 at the end of the year?
- Question : 13PS - What linear differential equation dy / dt = a( t )y is satisfied by y( t) = ecos t ?
- Question : 14PS - If the interest rate is a = 0.l per year in y' = ay, how many years does it take for your investment to be multiplied by e ? How many years to be multiplied by e2 ?
- Question : 15PS - Write the first four terms in the series for y = et2
- Question : 16PS - Find the derivative of Y ( t) = ( 1 + ! ) n. If n is large, this dY / dt is close to Y !n
- Question : 17PS - Suppose the exponent in y = eu(t) is u(t) = integral of a(t). What equation dy / dt = __ y does this solve ? If u(0) = 0 what is the starting value y(0) ?
- Question : 18PS - Applying this series to f(x) at x = 0 would give f + f' +
- Question : 19PS - (Computer or calculator, 2.xx is close enough) Find the time t when et = 10. The initial y(0) has increased by an order of magnitude-a factor of 10. The exact statement of the answer is t = . At what time t does et reach 100 ? -t2/2
- Question : 20PS - The most important curve in probability is the bell-shaped graph of e With a calculator or computer find this function at t = -2,-1,0,1,2. Sketch the graph of e-t212 from t = -oo tot = oo. It never goes below zero.
- Question : 21PS - Explain why Y1 = e(a+b+c)t is the same as Y2 = eatebtect_ They both start at y(0) = 1. They both solve what differential equation?
- Question : 22PS - For y' = y with a = l, Euler's first step chooses Y1 = (1 + .6.t)Y0. Backward Euler chooses Y1 = Y0/(1 -.6.t). Explain why 1 + .6.t is smaller than the exact et:,.t and 1/ (1 -.6.t) is larger than et:,.t. (Compare the series for 1 / (1 -x) with ex.) Note Section 3.5 presents an accurate Runge-Kutta method that captures three more terms of eaM than Euler. For dy / dt = ay here is the step to Yn+l : ( a2 .6.t2 a3 .6.t3 a4 .6.t4) Runge-Kutta for y 1 = ay Yn+l = l + a.6.t + 2-+ 6-+ ? Yn .
- Question : 1PS - y = Yoo = _. That constant y = y00 is a particular solution YpWhich Yn = ce-t combines with this steady state Yp to start from y(O) = 4? This question chose Yp + Yn to be y00+ transient (decaying to zero).
- Question : 2PS - For the same equation dy / dt = -y + 2, choose the null solution Yn that starts from y(O) = 4. Find the particular solution Yr that starts from y(O) = 0. This splitting chooses the two parts eaty(O) + integral of ea(t-s)q in equation (4).
- Question : 3PS - The equation dy/dt = -2y + 8 has two natural splittings Ys + YT = YN + YP: 1. Steady (ys = y00) + Transient (YT -+ 0). What are those parts if y(O) = 6? 2. (yfv = -2yN from YN(0) = 6) + (y; = -2yp + 8 starting from yp(0) = 0).
- Question : 4PS - All null solutions to u -2v = 0 have the form ( u, v) = ( c, __ ) . One particular solution to u -2v = 3 has the form (u, v) = (7, __ ). Every solution to u -2v = 3 has the form (7, __ ) + c(l, __ ). But also every solution has the form (3, __ ) + C(l, __ ) for C = c + 4.
- Question : 5PS - The equation dy/dt = 5 with y(O) = 2 is solved by y = __ . A natural splitting Yn(t) = _ and Yp(t) = _ comes from Yn = eaty(O) and Yp = J ea(t-s)5 ds. This small example has a = 0 (so ay is absent) and c = 0 (the source is q = 5e0t). When a= c we have "resonance." A factor twill appear in the solution y.
- Question : 6PS - For these equations starting at y(O) = 1, find Yn(t) and yp(t) and y(t) = Yn + Yp
- Question : 7PS - Find a linear differential equation that produces Yn(t) = e2t and yp(t) = 5(e8t - 1).
- Question : 8PS - Find a resonant equation ( a = c) that produces Yn ( t) = e2t and Yp ( t) = 3te2t.
- Question : 9PS - y' = 3y + e3t has Yn = e3ty(O). Find the resonant Yp with Yp(O) = 0. Problems 10-13 are about y' - ay = constant source q.
- Question : 10PS - Solve these linear equations in the form y = Yn + Yp with Yn = y(O)eat_ (a) y' - 4y = -8 (b) y' + 4y = 8 Which one has a steady state?
- Question : 11PS - Find a formula for y ( t) with y ( 0) = 1 and draw its graph. What is y00 ? (a) y' + 2y = 6 (b) y' + 2y = -6
- Question : 12PS - Write the equations in Problem 11 as Y' = -2Y with Y = y -y00
- Question : 13PS - If a drip feeds q = 0.3 grams per minute into your arm, and your body eliminates the drug at the rate 6y grams per minute, what is the steady state concentration y00 ? Then in= out and Yoo is constant. Write a differential equation for Y = y -y00 . Problems 14-18 are about y' - ay = step function H(t - T):
- Question : 14PS - Why is y00 the same for y' + y = H(t -2) and y' + y = H(t - 10)?
- Question : 15PS - Draw the ramp function that solves y' = H(t -T) with y(0) = 2.
- Question : 16PS - Find Yn(t) and Yp(t) as in equation (10), with step function inputs starting at T = 4. (a) y' -5y = 3H(t -4) (b) y' + y = 7H(t -4) (What is y00 ?)
- Question : 17PS - Suppose the step function turns on at T = 4 and off at T = 6. Then q(t) H(t -4) -H(t - 6). Starting from y(0) = 0, solve y' + 2y = q(t). What is y00?
- Question : 18PS - Suppose y' = H(t - 1) + H(t -2) + H(t -3), starting at y(0) = 0. Find y(t). Problems 19-25 are about delta functions and solutions toy' - ay = q 8(t - T).
- Question : 19PS - For all t > 0 find these integrals a(t), b(t), c(t) of point sources and graph b(t): t t t (a)J o(T -2) dT (b) J (o(T -2) -o(T -3)) dT (c) J o(T -2)o(T -3)dT 0 0 0
- Question : 20PS - Why are these answers reasonable? (They are all correct.) 00 00 00 (a) J eto(t)dt = l Cb) j (o(t))2dt= oo (c) J eT o(t -T)dT = et -00 -oo -oo
- Question : 21PS - The solution toy' = 2y + o(t -3) jumps up by lat t = 3. Before and after t = 3, the delta function is zero and y grows like e2t. Draw the graph of y(t) when (a) y(0) = 0 and (b) y(0) = 1. Write formulas for y(t) before and after t = 3.
- Question : 22PS - Solve these differential equations starting at y(0) = 2: (a) y' -y = o(t -2) (b) y' + y = o(t -2). (What is y00 ?)
- Question : 23PS - Solve dy/dt = H(t - 1) + o(t - 1) starting from y(0) = 0: jump and ramp.
- Question : 24PS - (My small favorite) What is the steady state y00 for y' = -y + o(t - 1) + H(t -3)?
- Question : 25PS - Which q and y(0) in y' -3y = q(t) produce the step solution y(t) = H(t - 1)?
- Question : 26PS - Solve these equations y' - ay = Qect as in (19), starting from y(O) = 2: (a) y' - y = 8e3t (b) y' + y = 8e-3t (What is Yoo ?) 29
- Question : 27PS - When c = 2.01 is very close to a = 2, solve y' - 2y = ect starting from y(O) = 1. By hand or by computer, draw the graph of y( t) : near resonance.
- Question : 28PS - When c = 2 is exactly equal to a = 2, solve y' - 2y = e2t starting from y(O) = 1. This is resonance as in equation (20). By hand or computer, draw the graph of y(t).
- Question : 29PS - Solve y' + 4y = 8e-4t + 20 starting from y(O) = 0. What is y00 ?
- Question : 30PS - The solution toy' -ay = ect didn't come from the main formula ( 4 ), but it could. Integrate e-asecs in (4) to reach the very particular solution (ect - eat)/(c - a).
- Question : 31PS - The easiest possible equation y' = 1 has resonance ! The solution y = t shows the factor t. What number is the growth rate a and also the exponent c in the source?
- Question : 32PS - Suppose you know two solutions y1 and Y2 to the equation y' - a(t)y = q(t). (a) Find a null solution toy' - a(t)y = 0. (b) Find all null solutions Yn
- Question : 33PS - Tum back to the first page of this Section 1.4. Without looking, can you write down a solution to y 1 - ay = q( t) for all four source functions q, H ( t), '5 ( t), ect ?
- Question : 34PS - Three of those sources in Problem 33 are actually the same, if you choose the right values for q and c and y(O). What are those values?
- Question : 35PS - What differential equations y 1 = ay + q( t) would be solved by y1 ( t) and y2 ( t) ? Jumps, ramps, corners-maybe harder than expected (math.mit.edu/dela/Pset1 .4).
- Question : 1PS - These steps lead again to the sinusoidal identity. This approach doesn't start with the usual formula cos ( wt -
- Question : 2PS - To express sin5t + cos5t as R cos (wt -
- Question : 3PS - To express 6 cos 2t + 8 sin 2t as R cos (2t -
- Question : 4PS - Integrate cos wt to find (sin wt)/w in this complex way. (i) dyreal/ dt= cos wt is the real part of dYcomplex/ dt= eiwt. (ii) Take the real part of the complex solution.
- Question : 5PS - The sinusoidal identity for A = 0 and B = -l says that -sin wt = R cos( wt -
- Question : 6PS - Why is the sinusoidal identity useless for the source q(t) = cos t + sin 2t?
- Question : 7PS - Write 2+3i as rei, so that 2_;3i = ?e-i. Then write y = eiwt / (2+3i) in polar form. Then find the real and imaginary parts of y. And also find those real and imaginary parts directly from (2 - 3i)eiwt / (2 - 3i) (2 + 3i).
- Question : 8PS - Write these functions A cos wt + B sin wt in the form R cos( wt -
- Question : 9PS - Solve dy / dt= 2y + 3 cost+ 4 sin t after recognizing a and w. Null solutions Ce2t.
- Question : 10PS - Find a particular solution to dy / dt= -y -cos 2t.
- Question : 11PS - What equation y 1 - ay = A cos wt + B sin wt is solved by y= 3 cos 2t + 4 sin 2t?
- Question : 12PS - The particular solution to y 1 = y + cost in Section 1.4 is Yp = et J e-s cos s ds. Look this up or integrate by parts, from s = 0 tot. Compare this Yp to formula (3).
- Question : 13PS - Find a solution y = M cos wt+ N sin wt toy' - 4y = cos 3t + sin 3t.
- Question : 14PS - Find the solution toy' - ay = A cos wt+ B sin wt starting from y(O) = 0.
- Question : 15PS - If a= 0 show that Mand Nin equation (3) still solve y' = A cos wt+ B sin wt. Problems 16-20 solve the complex equation y 1 - ay = Rei(wt-).
- Question : 16PS - Write down complex solutions Yp = Yeiwt to these three equations: (a) y' - 3y = 5e2it (b) y' = Rei(wt-
- Question : 17PS - Find complex solutions Zp = zeiwt to these complex equations: (a) z' + 4z = e8it (b) z' + 4iz = e8it (c) z' + 4iz = e8t
- Question : 18PS - Start with the real equation y 1 -ay= R cos ( wt-cp). Change to the complex equation z 1 - az = Rei(wt-
- Question : 19PS - What is the initial value Yp(O) of the particular solution Yp from Problem 18? If the desired initial value is y(O), how much of the null solution Yn = Ceat would you add to Yp ?
- Question : 20PS - Findtherealsolutiontoy'-2y =cos wtstartingfromy(O) = O,inthreesteps: Solve the complex equation z 1 - 2z = eiwt, take Yp = Re z, and add the null solution Yn =Ce2t with the right C.
- Question : 21PS - Find rand a to write each iw - a as reia. Then write 1/reia as Ge-ia. (a) v3i+l (b) v3i -1 (c) i - v3
- Question : 22PS - Use G and a from Problem 21 to solve (a)-(b)-(c). Then take the real part of each equation and the real part of each solution. (a) y' + y= eh!'3t (b) y' -y = eiv'3t (c) y' - \/3y = eit
- Question : 23PS - Solve y' - y = cos wt+ sin wt in three steps: real to complex, solve complex, take real part. This is an important example. (1) Find R and
- Question : 24PS - Solve y 1 - v'3y = cos t + sin t by the same three steps with a = v'3 and w = l.
- Question : 25PS - Challenge) Solve y 1 - ay = A cos wt + B sin wt in two ways. First, find R and
- Question : 26PS - We don't have resonance for y 1 - ay = Reiwt when a and w -:/- 0 are real. Why not? (Resonance appears when Yn = Ceat and Yp = Yect share the exponent a= c.) 27
- Question : 27PS - If you took the imaginary party= Im z of the complex solution to z 1 -az = Rei(wt-), what equation would y(t) solve? Answer first with
- Question : 28PS - SolveLdI/dt+RI(t) = V cos wtforthecurrentI(t) =ln+I p intheRLloop. 29
- Question : 29PS - With L = 0 and w = 0, that equation is Ohm's Law V = IR for direct current. 30 31 The complex impedance Z = R + iwL replaces R when L-:/- 0 and I(t) = Ie iwt. LdI/dt + RI(t) = (iwL + R)Jeiwt = Veiwt gives Z I= V. What is the magnitude IZI = IR+ iwLI? What is the phase angle in Z = 1Zlei0? Is the current II Ilarger or smaller because of L?
- Question : 30PS - Solve R- + -q(t) = V cos wt for the charge q(t) = Qn + Qp in the RC loop. dt C
- Question : 31PS - Why is the complex impedance now Z = R + i?C? Find its magnitude IZI. Note that mathematics prefers i = v=T, we are not conceding yet to j = v=T !
- Question : 1PS - Solve the equation dy / dt = y + l up to time t, starting from y(0) = 4.
- Question : 2PS - You have $1000 to invest at rate a = l = 100 %. Compare after one year the result of depositing y(0) = 1000 immediately with q = 0, or choosing y(0) = 0 and q= 1000/year to deposit continually during the year. In both cases dy/dt = y + q.
- Question : 3PS - If dy / dt = y - l, when does your original deposit y(0) =
- Question : 4PS - Solve dy = y + t2 from y(0) = 1 with increasing source term t2. dt
- Question : 5PS - Solve !? = y + et (resonance a = c !) from y(0) = 1 with exponential source et.
- Question : 6PS - Solve !? = y -t2 from an initial deposit y(0) = 1. The spending q(t) = -t2 is growing. When (if ever) does y( t) drop to zero? Solve !? = y -et from an initial deposit y(0) =
- Question : 7PS - Solve !? = y -et from an initial deposit y(0) = 1. This spending term -et grows at the same et rate as the initial deposit. When (if ever) does y drop to zero ?
- Question : 8PS - Solve dy = y -e2t from y(0) = 1. At what time Tis y(T) = 0 ?
- Question : 9PS - Which solution (y or Y) is eventually larger if y(O) = 0 and Y(0) = 0? dy dt = y + 2t or dY dt = 2Y +t.
- Question : 10PS - Compare the linear equation y 1 = y to the separable equation y 1 = y2 starting from y(O) = 1. Which solution y(t) must grow faster? It grows so fast that it blows up to y(T) = oo at what time T ?
- Question : 11PS - Y' = 2Y has a larger growth factor (because a = 2) than y' = y + q(t). What source q(t) would be needed to keep y(t) = Y(t) for all time? 12
- Question : 12PS - Starting from y(O) = Y(0) = 1, does y(t) or Y(t) eventually become larger? dy dt = 2y + et dY dt = y + e2t_
- Question : 13PS - What is the factor G(s,s) in zero time? Find G(s,oo) if a= -1 and if a= l.
- Question : 14PS - Explain the important statement after equation (13): The growth factor G(s, t) is the solution to y' = a(t)y + o(t - s). The source o(t - s) deposits $1 at time s.
- Question : 15PS - Now explain this meaning of G(s, t) when tis less than s. We go backwards in time. For t < s, G ( s, t) is the value at time t that will grow to equal 1 at time s. When t = 0, G(s, 0) is the "present value" of a promise to pay $1 at time s. If the interest rate is a = 0.1 = 10 % per year, what is the present value G(s, 0) of a million dollar inheritance promised in s = 10 years ? 16
- Question : 16PS - (a) What is the growth factor G(s, t) for the equation y' = (sin t)y + Q sin t? (b) What is the null solution Yn = G(O, t) toy'= (sin t)y when y(O) = 1? ( c) What is the particular solution Yp = J G ( s, t) Q sin s ds ?
- Question : 17PS - (a) What is the growth factor G(s, t) for the equation y' = y/(t + 1) + 10? (b) What is the null solution Yn = G(O, t) toy'= y/(t + 1) with y(O) = 1? t (c) What is the particular solution yp = lOJG(s,t)ds?
- Question : 18PS - Why is G(t, s) = 1/G(s, t)? Why is G(s, t) = G(s, S)G(S, t)?
- Question : 19PS - (recommended) If dy / dt = ay + qeiwt, with t in seconds and y in meters, what are the units for a and q and w ?
- Question : 20PS - The logistic equation dy/dt = ay - by2 often measures the time tin years (and y counts people). What are the units of a and b?
- Question : 21PS - Newton's Law is m d2y / dt 2 + ky = F. If the mass m is in grams, y is in meters, and t is in seconds, what are the units of the stiffness k and the force F ?
- Question : 22PS - Why is our favorite example y 1 = y + l very unsatisfactory dimensionally ? Solve it anyway starting from y(0) = -1 and from y(0) = 0.
- Question : 23PS - The difference equation Yn+l = cYn + Qn produces Y1 = cYo + Qo. Show that the next step produces Y2 = c2Y0 + cQ0 + Q1. After N steps, the solution formula for Y N is like the solution formula for y1 = ay + q(t). Exponentials of a change to powers of c, the null solution eaty(0) becomes cNY0. The particular solution t YN = cN-1Qo +
- Question : 24PS - Suppose a fungus doubles in size every day, and it weighs a pound after 10 days. If another fungus was twice as large at the start, would it weigh a pound in 5 days
- Question : 1PS - If y(0)= a/2b, the halfway point on the S-curve is at t= 0.Show that d = band a a l a y(t) = d t b = -b t . Sketch the curve from Y-oo= 0 to y00 = -.
- Question : 2PS - If the carrying capacity of the Earth is K = a/b = 14 billion people, what will be the population at the inflection point? What is dy / dt at that point? The actual population was 7.14 billion on January 1, 2014.
- Question : 3PS - Equation (18) must give the same formula for the solution y(t) as equation (16). If the right side of ( 18) is called R, we can solve that equation for y : ---+ y=--?(l + R
- Question : 4PS - Change the logistic equation to y' = y+ y2. Now the nonlinear term is positive, and cooperation of y with y promotes growth. Use z = l/y to find and solve a linear equation for z, starting from z(0) = y(0) = 1. Show that y(T) = oo when e-T = 1/2. Cooperation looks bad, the population will explode at t = T.
- Question : 5PS - The US population grew from 313,873,685 in 2012 to 316,128,839 in 2014. If it were following a logistic S-curve, what equations would give you a, b, din the formula (4)? Is the logistic equation reasonable and how to account for immigration?
- Question : 6PS - The Bernoulli equation y' = ay - byn has competition term byn. Introduce z = y1-n which matches the logistic case when n = 2. Follow equation (4) to show that z' = (n - l)(-az + b). Write z(t) as in (5)-(6). Then you have y(t).
- Question : 7PS - y' = y -y2 is solved by y(t) = 1/(de-t + 1). This is an S-curve when y(0) = 1/2 and d = 1. But show that y( t) is very different if y(0) > 1 or if y(0) < 0. If y(0) = 2 then d =
- Question : 8PS - (recommended) Show those 3 solutions to y' = y -y2 in one graph ! They start from y(0) = 1/2 and 2 and -1. The S-curve climbs from
- Question : 9PS - Graph f(y) = y -y2 to see the unstable steady state Y = 0 and the stable Y = 1. Then graph f(y) = y-y2 - 2/9 with harvesting h = 2/9. What are the steady states Y1 and Y2? The 3 regions in Problem 8 now have Z-curves above y = 2/3, S-curve sandwiched between 1/3 and 2/3, dropoff curves below y = 1/3. 10
- Question : 10PS - hat equation produces an S-curve climbing to y00 = K from Y-oo = L?
- Question : 11PS - y 1 = y -y2 -
- Question : 12PS - Solve the equation y 1 = -(y -
- Question : 13PS - With overharvesting, every curve y(t) drops to -oo. There are no steady states. Solve Y -Y2 - h = 0 (quadratic formula) to find only complex roots if 4h > 1. The solutions for h =
- Question : 14PS - With two partial fractions, this is my preferred way to find A= -1-, B = 1-r -s s-r PF2 1 1 1 ------=------+-----(y - r)(y - s) (y - r)(r - s) (y - s)(s - r) Check that equation : The common denominatoron the right is ( y - r) ( y - s) ( r - s). The numerator should cancel the r - s when you combine the two fractions. 1 1 . f . ABSeparate 2 -- and 2 -- mto two ractions -- + --.y1 y-y y-r y-s Note When y approaches r, the left side of PF2 has a blowup factor 1/(y -r). The other factor 1/(y - s) correctly approaches A = 1/(r - s). So the right side of PF2 needs the same blowup at y = r. The first term A/(y -r) fits the bill.
- Question : 15PS - The threshold equation is the logistic equation backward in time : dy 2 --= ay - by is the same as dt dy 2 -= -ay+by. dt 63 Now Y = 0 is the stable steady state. Y = a/b is the unstable state (why?). If y(0) is below the threshold a/b then y(t) -t 0 and the species will die out. Graph y(t) with y(0) < a/b (reverse S-curve). Then graph y(t) with y(0) > a/b.
- Question : 16PS - (Cubic nonlinearity) The equation y1 = y(l - y)(2 - y) has three steady states: Y = 0, 1, 2. By computing the derivative df / dy at y = 0, 1, 2, decide whether each of these states is stable or unstable. Draw the stability line for this equation, to show y(t) leaving the unstable Y's. Sketch a graph that shows y(t) starting from y(0) =
- Question : 17PS - (a) Find the steady states of the Gompertz equation dy/ dt = y(l - In y). (b) Show that z = In y satisfies the linear equation dz/ dt = 1 - z. (c) The solution z(t) = 1 + e-t(z(0) - 1) gives what formula for y(t) from y(0)?
- Question : 18PS - Decide stability or instability for the steady states of (a) dy/dt = 2(1 - y)(l - eY) (b) dy/dt = (1 - y2)(4 - y2)
- Question : 19PS - Stefan's Law of Radiation is dy/ dt = K(M4 -y4). It is unusualto see fourth powers. Find all real steady states and their stability. Starting from y(0) = M /2, sketch a graph of y(t).
- Question : 20PS - dy / dt = ay - y3 has how many steady states Y for a < 0 and then a > 0 ? Graph those values Y (a) to see a pitchfork bifurcation-new steady states suddenly appear as a passes zero. The graph of Y (a) looks like a pitchfork
- Question : 21PS - (Recommended) The equation dy / dt = sin y has infinitely many steady states. What are they and which ones are stable? Draw the stability line to show whether y(t) increases or decreases when y(0) is between two of the steady states.
- Question : 22PS - Change Problem 21 to dy/dt = (sin y)2
- Question : 23PS - (Research project) Find actual data on the US population in the years 1950, 1980, and 2010. What values of a, b, d in the solution formula (7) will fit these values? Is the formula accurate at 2000, and what population does it predict for 2020 and 2100? You could reset t = 0 to the year 1950 and rescale time so that t = 3 is 1980.
- Question : 24PS - If dy/dt= f(y), what is the limit y(oo) starting from each point y(0)? l?(y) --1-0- ---2?----
- Question : 25PS - (a)Draw a function f(y) so that y(t) approaches y(oo) = 3 from every y(0). (b)Draw f(y) sothaty(oo)= 4ify(0) > 0andy(oo) = -2ify(0) < 0.
- Question : 26PS - Which exponents n in dy / dt = yn produce blowup y(T) = oo in a finite time? You could separate the equation into dy/yn = dt and integrate from y(0) = 1.
- Question : 27PS - Find the steady states of dy / dt = y2 -y4 and decide whether they are stable, unstable, or one-sided stable. Draw a stability line to show the final value y( oo) from each initial value y(0).
- Question : 28PS - For an autonomous equation y 1 = f (y), why is it impossible for y( t) to be increasing at one time ti and decreasing at another time t2 ?
- Question : 1PS - Finally we can solve the example dy / dt = y2 in Section 1.1 of this book. y t Start from y(O) = 1. Then J ?; = J dt. Notice the limits on y and t. Find y(t).
- Question : 2PS - Start the same equation dy/dt = y2 from any value y(O). At what time t does the solution blow up? For which starting values y(O) does it never blow up? 3
- Question : 3PS - Solve dy/dt = a (t) y as a separable equation starting from y(O) = 1, by choosingf(y) = 1/y. This equation gave the growth factor G(O, t) in Section 1.6. 4
- Question : 4PS - Solve these separable equations starting from y(O) = 0: 5 6 dy (a) dt = ty (b)
- Question : 5PS - Solve d d y = a(t)y2 = a/(t? as a separable equation starting from y(O) = 1.t 1 y
- Question : 6PS - The equation !? = y + t is not separable or exact. But it is linear and y = __
- Question : 7PS - The equation ddy = '!j_ has the solution y = At for every constant A. Find this solution t t by separating f = 1/y from g = 1/t. Then integrate dy/y = dt/t. Where does the constant A come from ?
- Question : 8PS - F h. h b A . dy ct -ay . ? F h. A 1 h or w IC num er IS - = A an exact equation . or t 1s , so ve t e dt t+by equation by finding a suitable function F(y, t) + C(t).
- Question : 9PS - Find a function y ( t) different from y = t that has dy / dt = y2 / t2.
- Question : 10PS - These equations are separable after factoring the right hand sides : Solve dy and dt = yt + y + t + 1.
- Question : 11PS - These equations are linear and separable: Solve dy = (y + 4) cost and dy = yet.
- Question : 12PS - Solve these three separable equations starting from y(O) = 1 : dy ( a) -= -4ty dt (b) dy -= ty3 dt (c) dy (1 + t) dt = 4y
- Question : 13PS - (a) dy = -3t2 - 2y2 (b) dy 1+yety dt 4ty + 6y2 dt 2y + tety
- Question : 14PS - (a) dy 4t-y (b) dy 3t2 + 2y2 dt t-6y dt 4ty + 6y2
- Question : 15PS - Sh h dy y2 . b h . dy y . ow t at d = -- 1s exact ut t e same equation - = -- 1s not exact. t 2ty dt 2t Solve both equations. (This problem suggests that many equations become exact when multiplied by an integrating factor.)
- Question : 16PS - Exactness is really the condition to solve two equations with the same function H ( t, y) aH aH . aJ ag 8y = f(t, y) and 8t = -g(t, y) can be solved 1f at = -ay. Take the t derivative of aH / ay and the y derivative of a H / at to show that exactness is necessary. It is also sufficient to guarantee that a solution H will exist.
- Question : 17PS - The linear equation !; = aty + q is not exact or separable. Multiply by the integrating factor e-fat dt and solve the equation starting from y(O).
- Question : 18PS - (y is missing) Solve these differential equations for v = y I with v ( 0) = 1. Thensolve for y with y(0) = 0. (a) y11+y'=0 (b) 2ty" -y' = 0.
- Question : 19PS - Bothy and tare missing in y" = (y ')2. Set v = y I and go two ways: dv dy I. (y missing) Solve -d = v2 for v(t) and then -= v(t) t dt II. ( t missing) with y(0) = 0, y '(0) = 1. dv dy Solve v dy = v2 for v(y) and then dt = v(y)with y(0) = 0, y'(0) = 1.
- Question : 20PS - An autonomous equation y' = f(y) has no terms that contain t (t is missing). Explain why every autonomous equation is separable. A non-autonomous equation could be separable or not. For a linear equation we usually say LTI (linear timeinvariant ) when it is autonomous: coefficients are constant, not varying with t.
- Question : 21PS - my 11 + ky = 0 is a highly important LTI equation. Two solutions are cos wt and sin wt when w2 = k/m. Solve differently by reducing to a first order equation for y1 = dy / dt = v with y 11 = v dv / dy as above : mv !? + ky = 0 integrates to ?mv2 + ?ky2 = constant E. For a mass on a spring, kinetic energy
- Question : 22PS - my 11 + k sin y = 0 is the nonlinear oscillation equation: not so simple. Reduce to a first order equation as in Problem 21 : dv k . 0 . 1 2 k mv dy + smy = mtegrates to 2mv - cosy= constant E. With v = dy / dt what impossible integral is needed for this first order separableequation? Actually that integral gives the period of a nonlinear pendulum-thisintegral is extremely important and well studied even if impossible

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