Math: HW# solutions Andy Royston October, 5 8..4 Integrate each side from to t: t d x dt dt dx dx (t) dt dt () g k e kt t t ge kt dt g k ( e kt ). () Since the object starts from rest, dx dx () v(). Now integrate v(t) (t) to get the distance dt dt traveled: t v(t )dt d(t) x(t) x() g k (t + k e kt ) t t g ( e kt )dt k g k t g k ( e kt ). () To look at the behavior for small t, Taylor expand and keep only the lowest order terms in t: or d(t) g k t + g ( ( kt + ) k k t O(t 3 )) g k t g k (kt k t + O(t 3 )) gt O(t 3 ), (3) d(t) gt, t small. (4) So the motion is approximately free-fall for small t. For large t, we have e kt. Therefore, the dominating term in the velocity is the constant: v(t) g, for t large. (5) k
8..7 We have F dp/dt, where F is a constant. Integrate both sides from to t: F t t dp dt dt p(t) p() mv(t) p(). (6) v(t) /c We could write out p() in terms of v(), but it s just a constant anyway. Now lets solve the above for v(t). ( v c )(F t + p ) m v v c (m c + (F tp ) ) (F t + p ), v(t) c F t + p m c + (F t + p ). (7) (I am using the common notation q() q for any quantity q). Let us look at this result in the limit t. Rewrite this as v(t) c F t + p (F t + p ) + m c (F t+p ) + m c /(F t + p ). (8) It is clear that this expression goes to in the limit since the fraction in the denominator goes to. Therefore v(t) lim, or lim v(t) c. (9) t c t Now we are asked to find the distance traveled, d(t), if the electron starts from rest. If the electron starts from rest, then v p. Hence, d(t) t v(t )dt c c F t F t dt m c + F t m c + F t t c F t F t dt m c + F t c F ( m c + F t mc). () 8.. x y dx + y x dy xdx ydy. () x y Now integrate each side. Here, it is better to do indefinite integrals and add a constant to one side. This is because we are not given initial data (such as y(x ) y ); rather, we are given the value at a pair of points. Then, xdx x x ydy y + c y + c x + y c. ()
.5.5 -.75-.5-.5.5.5.75 - -.5.5 -.5 -.5 - - Figure : Problem 8... The slope field and some solution curves. The values of c chosen for the curves ranged between and. See Figure for a plot of the slope field and some of the solution curves. Note that the result only gives real curves for x and y This implies c. We are told that the solution passes through (x, y ) (/, /). Plugging these values into the above equation: Hence, the solution is 3/4 3 c. (3) x + y 3. (4) One could solve this for y to get y as a function of x, but this is unnecessary. 8..4 ( + y )dx + xydy See Figure. We have (x, y ) (5, ), so Hence, dx x ydy + y + c ln x ln ( + y ) + c. (5) ln 5 ln + c c ln 5. (6) 3
3-3 - - 3 - - -3 3-3 - - 3 - - -3 Figure : Problem 8..4. Here the values of c were chosen between 3 and 3. ln x ln ( + y ) + ln 5 x 5 + y. (7) 8..8 dy dx + xy dy y xdx + c y x + c, or y x c. (8) See Figure 3. Setting (x, y) (, ) yields c 3. Thus, y(x) x 3. (9) 8..3 Observe that the differential equation x y dx + y x dy is satisfied when y ± for all x. This is because dy if y is a constant. These solutions can be guessed from looking at the slope field. Note that in the family of solutions we found, setting y ± would imply x c, () a contradiction since c is a constant. Hence, these are singular solutions to the non-linear diff. eq. 8..8 The differential equation is dn dt N, () 4
- - - - - - - - Figure 3: Problem 8..8. The c values were again taken from 3 to 3. For c 3,...,, each curve has two poles, where it blows up to ±. c,, 3 correspond to the bounded curves. and we are given the initial conditions N(t ). Let us first solve this by separation of variables. dn N dt + c N t + c N(t) 4 (t + c). () The solution which satisfies N() is c. Hence, we obtain N(t) t /4. But this solution is physically unreasonable. If the bacteria grows at a rate proportional to the square root of the amount, and the amount is initially zero, it will not grow. The correct solution is N(t), t. Note that this is a solution to the differential equation, though it is not in the family we obtained. We essentially lost it when we divided by N. 8.. We have dn dt kn dn N kdt + c ln N kt + c N(t) Ae kt. (3) Using the initial condition, N() A N, we find N(t) N e kt. (4) 5
8..6 m k m dv dt mg kv dv (g kv/m)dt dv g kv/m dt + c k ln (g kv/m) t + c ln (g kv/m) (t + c) g kv/m Ae kt/m m v(t) mg k m k Ae kt/m. (5) Now, we are told that the object falls from rest, so Hence, the solution is v() mg k m A A g. (6) k v(t) mg k ( e kt/m ). (7) The limiting value of v as t is the terminal speed. Since e kt/m, we have v term mg/k. However, assuming a terminal speed exists, we can also find it directly from the differential equation. If v v term, a constant, then we must have dv/dt. Plugging this into the differential equation, mg kv term v term mg/k. (8) a Suppose a kg person parachutes from a plane. A reasonable value of v term might be m/s. Then, in SI units, ()(9.8) k k 9, 8kg/s. (9) b Note that we may write the solution for v(t) as v(t) v term ( e kt/m ). We wish to solve for the time such that v(t).99v term :.99v term v term ( e kt/m ) e kt/m. t m k ln. 4.65m k. (3) 8..3 We need to find the equation for the tangents of the orthogonal trajectories. Start by taking the differential of the given equation, which represents circles centered at the origin, xdx + ydy dy dx x y. (3) This gives the tangents of the slope field. Note that the constant dropped out. Since the constant tells us which curve (circle) we are looking at, this means that our expression above for the tangent lines is correct for any curve, ie. for any value of (x, y). The tangents of the orthogonal trajectories are the negative reciprocals: 6
.5.5 -.5 - -.5.5.5 -.5 - -.5 Figure 4: Problem 8..3. Circles are original curves; straight line are orthogonal trajectories. 4 3-3 - - 3 - - Figure 5: Problem 8..35. Original curves in black; orthogonal trajectories in red. dy o.t. dx y o.t. x dyo.t. dx y o.t. x + c ln y o.t. ln x + c y o.t. e c x Ax. (3) This makes sense. The orthogonal trajectories of circles centered about the origin are straight lines passing through the origin. See Figure 4. 8..35 Do the same thing as in the last problem. (y )dy xdx dy dx 7 x y. (33)
Again, the constant drops out. Hence, dy o.t dx y o.t. x dy o.t. dx y o.t. x + c ln ( y c.t.) ln x + c y o.t. e ln x e c A/x y o.t. A/x. (34) See Figure 5. 8.3.5 The differential equation has the form y (x) + y(x) cos x. (35) cos x We need to compute I dx sec x. This is a tricky integral to do by hand, but I get a rather unwieldy expression if I use Mathematica. So, here s how we can do it by hand. Think about it the other way around. What function might give us / cos x when we take its derivative. Consider the function ln (tan z). Its derivative is d ln (tan z) dz tan z (sec z) sin z cos z sin (z). (36) We want / cos x, but remember that cosine and sine are just shifted by π/. Therefore, consider setting z x + π/, or z x/ + π/4. Then Hence, d dx ln (tan (x/ + π/4)) tan (x/ + π/4) (sec (x/ + π/4)) ( ) sin (x/ + π/4) cos (x/ + π/4) sin (x + π/) cos x. (37) I dx cos x ln (tan (x/ + π/4)). (38) For the general solution, we still need to compute the integral Qe I : dxqe I dx cos x tan (x/ + π/4). (39) Let us change variables to u x/ + π/4, dx du. Then cos x cos (u π/) sin u sin u cos u. (4) 8
Thus, dx cos x tan (x/ + π/4) 4 du sin u cos u sin u cos u 4 du(sin u) du( cos u) (u sin (u)) u sin (u) x + π/ sin (x + π/) x cos x + π/. (4) The general solution is given by y e I ( Qe I + c). Hence, ( ) y(x) exp ( ln (tan (x/ + π/4))) x cos x + π/ + c ) cot (x/ + π/4) (x cos x + c. (4) When I plug the original differential equation into Mathematica, I get the solution (x cos x + c[])(cos (x/) sin (x/)) y cp (x). (43) cos (x/) + sin (x/) This is equivalent to my solution, as we can see by the following calculation: cot (x/ + π/4) cos (x/ + π/4) sin (x/ + π/4) cos (x/) cos (π/4) sin (x/) sin (π/4) sin (x/) cos (π/4) + cos (x/) sin (π/4). (44) Using sin (π/4) cos (π/4) ( / ), these factors cancel in the numerator and denominator. Hence, cot (x/ + π/4) cos (x/) sin (x/) cos (x/) + sin (x/). (45) It follows that our results are the same, with my c being Mathematica s c[]. 8.3.3 Note the hint at the bottom of page 43; we will solve for x as a function of y. dx + (x e y )dy dx dy (x ey ) x + x e y. (46) So P (y) and therefore, I dyp y. Meanwhile dyq(y)e I dye y ey. (47) Therefore, x(y) e y ( ey + c) ey + ce y. (48) This is exactly the form of the solution that Mathematica gives me. 9
8.3. See example of the text, pages 4 43. Suppose polonium decays (into something else) with decay constant λ 3. Polonium is being created at a rate equal to the rate at which radon is decaying, but it is being destroyed at it s decay rate. Therefore, the differential equation governing the quantity of polonium is dn 3 λ N λ 3 N 3 N 3 + λ 3 N 3 λ N (t), (49) dt where the right hand side of this equation has been solved for in the example and is given by λ N (t) λ λ N λ λ (e λ t e λ t ) A(e λ t e λ t ). (5) This solution assumes λ λ. We will further assume λ 3 λ and λ 3 λ. Also, I ve introduced the constant A to avoid cluttered formulae in what follows. Returning to the diff. eq. for N 3 (t), we see that P λ 3 is a constant, so I λ 3 t. Then dtq(t)e I A dt(e (λ 3 λ )t e (λ 3 λ )t ) A e(λ 3 λ )t λ 3 λ A e(λ3 λ)t λ 3 λ. (5) Then, N 3 (t) e I ( Qe I A + c) e λt A e λt + ce λ3t. (5) λ 3 λ λ 3 λ To fix c, we have the condition that there is no polonium at t : N 3 (). Solving this for c, Hence, we find A A ( ) + c c A λ 3 λ λ 3 λ λ 3 λ λ 3 λ A(λ λ ) (λ 3 λ )(λ 3 λ ) N λ λ (λ 3 λ )(λ 3 λ ). (53) N 3 (t) If we define N λ λ (λ λ )(λ 3 λ ) e λ t N λ λ (λ λ )(λ 3 λ ) e λ t + then this can be written in a more symmetric fashion: N λ λ (λ 3 λ )(λ 3 λ ) e λ 3t. (54) λ ij λ i λ j, (55) N 3 (t) N λ λ λ λ 3 λ 3 (λ 3 e λ t + λ 3 e λ t + λ e λ 3t ). (56)
4 3-4 -3 - - 3 4 - - -3-4 Figure 6: Problem 8.3.. Original curves in black; orthogonal trajectories in red. 8.3. Take the differential to get an equation for the tangent lines of the slope field: dx dy + ce y dy ( + ce y )dy dy dx + ce. (57) y Note that the constant c does not drop out in this case and we need to eliminate it. The constant c takes different values on different curves in the family, but we want a equation for the tangent lines that is valid at any point (x, y). This can be done by solving the original equation for the constant c in terms of x, y. x y + + ce y c e y (x y ). (58) Now we substitute this value into the expression for the tangent lines: dy dx + e y (x y )e y x y. (59) Then, the tangent lines to the orthogonal trajectories are dy o.t. dx y o.t. x y o.t. y o.t. x. (6) This is a first order linear equation. We have P, so I x. Then dxq(x)e I dxxe x xe x dxe x xe x + e x where I did an integration by parts. Hence, e x (x + ), (6) See Figure 6. y o.t. (x) e x (e x (x + ) + c ) x + + c e x. (6)
8.4. We have This is Bernoulli equation with n /. Let z y. Then, y + x y x/3 y. (63) z dz dx dz dy dy dx y y z y Replacing y with z, the differential equation becomes y zz. (64) zz + x z x 3/ z z + x z x3/, (65) which is a linear, first-order equation. We have P (x) /x, so I P ln x. Then Q(x)e I dxx 3/ x dxx x 3 /3. (66) Hence, the solution for z is and thus z(x) ( x3 + c), (67) x 3 8.5. Using the notation D d/dx, this equation has the form y(x) z(x) x (x3 3 + c). (68) (D + D )y (D + )(D )y. (69) The roots of the corresponding polynomial are r,. Hence, the general solution is y(x) c e x + c e x. (7)