# Linear Equations and Integrating Factors

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1. For y — by — 0 an integrating factor is e J ocix = e ox so that — [e oxy dx L J —oo < x < oo. There is no transient term. 2. For y1 + 2y — 0 an integrating factor is eJ 2 dx = e2x so that —oo < x < oo. The transient term is ce 2x. A dx <-2xy = 0 and y = ceox for = 0 and y — ce~2x for d 3. For y' + y — eix an integrating factor is eJ dx = ex so that — [e'Ty] -- and y - ie3* + ce~x for dx * 4 —oo < x < oo. The transient term is ce x. 4. For y' -f- Ay = 4 an integrating factor is eJ 4 dx == e4x so that [e4x dx L for — oo < x < oo. The transient term is ce~4x. V = |e4x and y = A + ce -4x 49 www.Elsolucionario.Net Exercises 2.3 Linear Equations 5. For y'+3x2y = x2 an integrating factor is eJ ,ix dx — e*3 so that dx -3 e y = x2ef3 and y = |+o. For — oo < x < oc. The transient term is ce—X 6. For y' + 2xy = x:i an integrating factor is eJ 2xdx — ex so that -r~dx t 9 O \x? — ^ + CC~X~ for — oo < x < oc. The transient term is ce~x~. J * e y = £3e* and .. = 11 d 1 1 7. For y' + — y = —~ an integrating factor is — x so that — [xy] = — and y = — In x x xL dx x x for 0 < x < oo. The entire solution is transient. 8 . For y' — 2y = x2 + o an integrating factor is e / 2dx = e 2x so that y - fe 2xy (JLJL and y = —^x2 - - j + ce2x for — oo < x < oc. There is no transient term. = x V 2* + O f' 1 1 d 9. For i f ---y = x sin x an integrating factor is e~JO/;r)rf3: = — so that — X •// CLX y — cx — x cos x for 0 < x < oo. There is no transient term. 2 3 x ' x for 0 < x < oo. The transient term is cx~2. 1 1 - y x . = sin a: 10. For i/ + —y = — an integrating factor is eJ (2/a:)rfx = x2 so that x2y| - dx and y = § + c:: t * t ri'r dx L * J z 4 dt 11. For y' + — y = :r2 — 1 an integrating factor is eJ (4i*)<lx = x 4 so that — xAy x ' dx 1 ■ y — — ix + cx-4 for 0 < x < oo. The transient term is cx~4. = x6 — x4 &: 12. For y'- x d y = x an integrating factor is e~Nx,/(1+x^dx = (x+l)e-a; so that — (x + 1 )e~xy (1 + x) dx l {^X | 3 x(x + l)e~x and y = —x — —— — H---— for —1 < x < oo. There is no transient term. V ' J x + 1 x + 1 ( 2 \ €>x ■ d H — ) y = —? An integrating factor is e-l t1+(2/*)]^ _ X2ex so \x2exy x j xz dx L ' J 2x e a: 1 ex c.E~x 2 x2 + t-2 for 0 < x < oo. The transient term is ce x2 14. For y' + (\ + ~') ?/ = —e x sin 2.X an integrating factor is = xex so that [xexy_ x x 1 ce sin 2x and y — — r—e x cos 2x + 2x x for 0 < x < oo. The entire solution is transient. 15. For ~ — -x = 4yD an integrating factor is e J(4^ dy — elny 4 = y 4 so that -7- \y 4x = Ay a:.: dy y rht fn " ' dy l J x = 2y 6 + cy4 for 0 < y < 00. There is no transient term. 50 www.Elsolucionario.Net Exercises 2.3 Linear Equations ^ + -x = ev an integrating factor is eJ^'2^ dy — y2 so that -7“ dy y ' dy 2 2 . C c r = -- H— s-e2' + for 0 < y < 00. The transient term is . Y y2 ir ' yz r d For y+(tan.X)?Y = sec a; an integrating factor is eJ tan x (lx = sec a* so that — [(sec.*) y\ = sec2x and dx j = sin x + c cos x for —tt/2 < x < tt/2. There is no transient term. For 2/ + (cot x)y - sec2x csc x an integrating factor is eJ cot x dx = eln-smxl = sin a; so that d - o [(sin x) y) — sec x and y = sec a; + ccsc x for 0 < x < tt/2. There is no transient term. T -1- 9 0'T.P~X r fl. For yf + -— ~ y = -—-— an integrating factor is eJ [(x+2)/(x+^)]dx — (x + l)ex. So — \(x + l)exy] = x + 1 ' x + 1 dx x2 c 2.R and u = ---- e x ---- - e x for — 1 < x < 00. The entire solution is transient. X + 1 x + 1 For y'-\--y - ^ an integrating factor is eJiA/(x+2)\dx — (x + 2)4 so that ~ [(a; + 2)4yl = x + 2 (x + 2)2 0 v J dx Lv 1 ~v.X + 2)2 and y - ^(a; + 2)_1 + c(x + 2)~4 for —2 < x < o o . The entire solution is transient, o dv For — + r see 6 — cos 9 an integrating factor is eJ s<]c()d0 = ein |seen-tan_ sec q _|_ tan/9 so that dt) ~ [(sec 9 + tan 6)r] — 1 + sin 9 and (sec 9 + tan 9)r = 9 — cos 9 + c for —tt/2 < 9 < 7t/2 . Id For + (21 — 1)P — 4t — 2 an integrating factor is g/(2t-1)rff = so that dt dt ■it — 2)et and P = 2 + cei_r for —oc < t < 00. The transient term is ce*-t . For 1/+ ( 3 + — ^ y = ---an integrating factor is e-/ [•i+(1/T)i,il: = xc*x so that — [xe3:ryj - 1 and V xJ x dx L ' J o, > .7 = e H-----for 0 < x < oo. The entire solution is transient. X For 1/ + ^ y = £JlJ: an integrating factor is e/i2/ ^ '-1)]*' = -— 1 so that — x* - 1 x - 1 x + 1 dx and (:/: — 1)y = x(x + 1) + c(x -f 1) for — 1 < x < 1. For y' + - y = — ex an integrating factor is e f^^:)dx — x so that ~ \xy\ - ex and y = —ex + — x x dx ' ’ x x 1 2 — e :or 0 < x < oc. If y(l) = 2 then c = 2 — e and y = —ex H---- -. X x For ~ x = 2y an integrating factor is e~ }(l/y)dy _ i so that — ^ dy y y dy ° f - l P x — 1 X + 1 . = 1 —X y - 2 and x = 2y2 + ay 49 for 0 < y < 00. If y(1) = 5 then c = —49/5 and x ■ 2y2— — y. 5 For f-j- + y I = T~ an integrating factor is eJ dt = eRt/L So that — \eRt'L i] — —eRt/L and dt L L dt 1 1 L 51 www.Elsolucionario.Net Exercises 2.3 Linear Equations E R dT i = ^ - + ce Rt/L for —oc < t < oc. If v'(0) = tQ then c = i o — E/R and i = — + ( i\$ — — 'j R R. \ RJ E R d r - i 28. For —-- kT = —Tmk an integrating factor is k^dt = e kt so that 4- fe kiT] - —Trnke kt an: (it dt T - Tm. + cekt for —oo < t < oo. If T(0) - To then c = 7q — Tm and T = Tm + (To — Tm)ekt. 29. For ?/ H-- —rV = —— r an integrating factor is = x + 1 so that ~-\(x + l)y| = x +1 x +1 dx X X c lnx and y = ——— In a,’ --- —- H---— for 0 < x < oc. If y(l) = 10 then c = 21 and &- + 1 x + 1 x + 1 X _ x 21 y = ---- hi x - x + 1 x + 1 ' x + 1 30. For i/+(tana)y = cos2x an integrating factor is e-ltanx<ir = ein|secx| — sec x so that — [(seex) y] = f.AtJU cosx and y = sin x cos x + a cos x for —tt/2 < x < tt/2. If y( 0) = —1 then c = —1 and y = sin x cos x — cosx. 31. For y' + 2y = f(x) an integrating factor is e2x so that e2x = { ie2x + ci, 0 < x < 3 W x > 3. If y(0) - 0 then c\ = —3/2 and for continuity we must have C2 — |e6 — | so that | i( l -e~2x), 0 < x < 3 V [ |(e6 - l)e~2a\ x > 3. 32. For y' -f y = f(x) mi integrating factor is ex so that ex + ci, 0 < x < 1 —ex + C2, x > 1. If y(0) = 1 then cj. = 0 and for continuity wc must have C2 = 2e so that 1 * - - yex = 1. 0 < x < 1 y = 2e1~x - 1, x > 1. 33. For y' + 2xy = /(x) an integrating factor is ex so that 2 f \ex~+ c\, 0 < x < 1 ye: = { 02, X > 1. If y(0) = 2 then cj = 3/2 and for continuity we must have C'2 = + 5 so that 52 www.Elsolucionario.Net Exercises 2.3 Linear Equations 4 . For , 2x y + T T * y v X i + fe - L, (he +1) 0 < x < 1 x > 1. 1 + X2 ' —x 1 + X2 ' 'ii integrating factor is 1 + x2 so that 0 < x < 1 x > 1, -1 — (l + x2) y = \x? + ci, 0 < x < 1 ~ lx 2 + C2, X > 1. Y(0) = 0 then ci = 0 and for continuity wc must have C2 = 1 so that r i i y - 2 2(1 + x2) ’ 0 < x < 1 [ 2 (1 + x2) 2 1: . *.Ve first solve the initial-value problem y' + 2y = 4a:. Y(0) = 3 on the interval j. 1]. The integrating factor is cJ 2dx = e2x. So 4-[e2xy} = ixe2x dx i2xy - j 4xe2xdx - 2xe2x — e2x + C\ - 2 x y = 2x — 1 + c\e Vsing the initial condition, we find y(0) = — 1 + c\ = 3, so c\ = 4 and , = 2x — 1 + 4e-2x, 0 < x < 1. Now, since y(l) = 2 — 1 + 4e~2 = 1 + 4e~2, -re solve the initial-value problem y' — (2fx)y = 4a;, y( 1) = 1 + 4e-2 on the interval (1. Oo). The integrating factor is eJ(-‘2/x)eix - c-2in* _ x-2^ s0 d . I _o 4 - [* ti= 4 x X - = - -2 f 4 * y = I ~ dx = 4 In a: + 7 a: y = Ax2 In x + C'yx2. C2 We use In a; instead of In |a;| because x >1.) Using the initial condition wc find y(l) = — l+4<? 2, ' ij = 4:r;2 In a: + (1 + 4e_2).X2, x > 1. Thus, the solution of the original initial-value problem is 53 www.Elsolucionario.Net Exercises 2.3 Linear Equations ( 2x — 1+ 4e~2x. 0 < x < 1 ^ \ 4.R2 In x + (1 + 4e_2)a;2. X > 1. See Problem 42 in this section. X ___ 36. For y' + exy = 1 an integrating factor is ee . Thus d r - ,.X , .X f X t — [( ' y\= e and e y = / e dt + c. Dx ' Jo From 2/(0) = 1 we get a = e, so y = e-6* Jq ee dt + e1-^ . When y' + exy = 0 we can separate variables and integrate: — = —ex dx and In lyj = — ex + c. V Thus y = c\e~^. Prom y(0) = 1 we get ci = e, so y = e1-e . When if + exy = ex we can see b,y inspection that y = 1 is a solution. 37. An integrating factor for y' — 2xy = 1 is e~x . Thus 2 rx ,2 \Fir e~x y = / e di = ^~- erf(z) + c */ 0 a y = ~~ex crf(x) + cex . & From y(l) = (v/7r/2)e erf(l) + ce = 1 we get c = e-1 — ^ crf(l). The solution of the initial-va/: problem is y = ^e^erffa) + (e~l - ^ e r f ^ e * 2 = ex -1 -r ex~(erf(x) — erf(l)). Z 38. We want 4 to be a critical point, so we use y' = 4 — y. 39. (a) All solutions of the form y = x°ex — x4ex 4- cx4 satisfy the initial condition. In this cafsince 4/x is discontinuous at x = 0, the hypotheses of Theorem 1.2.1 are not satisfied and t: initial-value problem does not have a unique solution. (b) The differential equation has no solution satisfying y(0) = (jq . Yo > 0. (c) In this case, since .-eo > 0, Theorem 1.2.1 applies and the initial-value problem has a unic solution given by y = x^ex — x4ex + c.R4 where c : vq/xq — XQex° + ex°. 40. On the interval (—3,3) the integrating factor is ej x d x / ( x 2—9) _ e - f x d x / ( 9 - x 2) _ g £ ln ( 9 - x 2) _ ^ /g _ ^.2 54 www.Elsolucionario.Net Exercises 2.3 Linear Equations and so d dx v'o" x2y - 0 and y = V9 —a:.2 We want the general solution to be y = 3x — 5 4- ce x. (Rather than e x, any function that approaches 0 as x —>• oo could be used.) Differentiating we get y' = 3 — ce-* = 3 — (y — 3a: + 5) = —y -f- 3x — 2, so the differential equation y1 + y = 3x — 2 has solutions asymptotic to the line y = 3.X — 5. The left-hand derivative of the function at x = 1 is 1/e and the right-hand derivative at x = 1 is 1 — 1/e. Thus, y is not differentiable at x = 1. (a) Differentiating yc = c/a:3 we get 3c y'c X41 so a differential equation with general solution yc = e/.R3 is xy' + 3y = 0. Now xy'p -f 3yp = x(3x2) + 3(.X3) = 6a:3 so a differential equation with general solution y — c/x3 + z3 is xy' -j- 3y = 6a:3. This will be a general solution on (0, oo). (b) Since y( 1) = l 3 — l / l 3 = 0, an initial condition is y( 1) = 0. Since 1/(1) = l 3 + 2/13 = 3, an initial condition is y(l) = 3. In each case the interval of definition is (0, oc). The initial-value problem xy' + 3y = 6a:3, y(0) = 0 has solution y = a:3 for —oo < x < oo. In the figure the lower curve is the graph of y(x) — a;3 — 1 /a:3.While the upper curve is the graph of y = x3 — 2/a:3. (c) The first two initial-value problems in part (b) are not unique. For example, setting 1/(2) = 23 - 1/23 = 63/8, we see that y(2) = 63/8 is also an initial condition leading to the solution y = a:3 — 1/a:3. Sincc ef p(x)dx+c = ec€f P(z)dx _ CieJ P{x)dX' we wou](j have cie-f p(x)dxy = C2 + J ^ ^ m d x and e^p^ dxy — cz + J eJ p^ cla: f(x) dx, which is the same as (6) in the text. We see by inspection that y = 0 is a solution. The solution of the first equation is x = eie_Al*. From a;(0) = x q we obtain ci = x q and so x — X Q e~ X lt . The second equation then becomes dy 4r = c — A-2y dt 55 www.Elsolucionario.Net Exercises 2.3 Linear Equations which is linear. An integrating factor is e*2*. Thus y [ex'2ty] = xoAie = aroAie^2 ^ Cot* M y = ^ Ai e(A2-Ai)f + ^ A2 - Ai .X'oAi . E“Ajt_|_c g A'2<_ A2 -Al From y(0) = yo we obtain C2 = (yoAa — yoAi — £oA])/(A2 — Ai). The solution is _ sqAi e_Alt + ypA2 - ypA] - rcpA] ^_Aa/, A2 — Ai A2 — Aj d.E 1 47. Writing the differential equation as —— h jE7 = Q we see that an integrating factor is et/Rl do J- i-O Then A[el:/RCE} = 0 dt ef!RCE = c E = ce~t/RC. From E(4) = ce~4/ Rc = Eq w c find c: = Eoe^Rc. Thus, the solution of the initial-value problem E = E0e^RCe ^ RC = E^ - ^ RC. 48. (a) An integrating factor for y' — 2xy = —1 is e~x2. Thus d r _r2 2 " v\ = ~e ypK e x y = — J e 1 dt = — erf (z) + c. From y(0) = v/tt/2, and noting that crf(0) = 0, we get c = a/ tt/ 2. Thus V = e’ ! OrfM + I f ) = T f - el{(x» = X e' 2 erfo(a:)' (b) Using a CAS wc find y(2) 0.226339. 49. (a) An integrating factor for , 2 10sin:r y + - y = a; .T 56 www.Elsolucionario.Net Exercises 2.3 Linear Equations is x2. Thus o „. Rx sin t , x y = 10 ---dt + c J Jo t y = 10# 2Si(;r) + ex 2. Prom y(l) = 0 we get c = — 10Si(l). Thus y = 10.X“2Si(#) - 10x_2Si(l) = 10a.R2(Si(ir) - Si(l)). ( b ) y r x > -1 -2 -3 -4 “5 (c) From the graph in part (b) we see that the absolute maximum occurs around x — 1.7. Using the root-finding capability of a CAS and solving yf(x) = 0 for x we sec that the absolute maximum is (1.688,1.742). So y = 7*x)mUsing 5(0) — 0 and y(0) = c\ — 5 we have y = 2 1 rX . 9 ( a) The integrating factor for y' — (sinx2)y = 0 is e~Jo . Then y = CleJosint2dt 57 www.Elsolucionario.Net Exercises 2.3 Linear Equations (c) From the graph wc see that as x —► oc, y(x) oscillates with decreasing amplitudes approaching 9.35672. Since limx-^oo 5S(x) = \ , linij;-*^ y(x) — 9.357, and since lima;_i._oc S(x) = —^ , lim ^-^c y(x) = 5e-v/^/8 & 2.672. (d) From the graph in part (b) we see that the absolute maximum occurs around x = 1.7 and tlicabsolute minimum occurs around x = —1.8. Using the root-finding capability of a CAS anc solving y'(x) = 0 for we see that the absolute maximum is (1.772,12.235) and the absolute minimum is (—1.772,2.044). 1. Let M = 2x — 1 and N = 3y+7 so that My = 0 = Nx. From fx = 2x— 1 we ob