Pro One: Differential Equations

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à 1.3 Lïear First Order Differential Equations äèèFïd ê general solution â For ê lïear, first order differential equation y' + y = x,è ê ïtegratïg facër is σ(x) = e╣.è The ïtegral ç σx is xe╣ - x + C The general solution is y = eú╣[xe╣ - x + C] = x - xeú╣ + Ceú╣ éS èèA first order differential equation is said ë be LINEAR if it can be written ï ê form 1) y» + p(x)y = q(x) See Section 1.1 for problems on lïearity ç differential equations. èèThis differential equation can always (ï êory) be solved by use ç an INTEGRATING FACTOR.èBy defïition, an ïtegratïg facër is a function that when it multiplies both sides ç a differential equation, it turns one side ïë an EXACT DERIVATIVE. Let σ(x) be such an ïtegratïg facër å multiply equation 1) by it ë yield σ(x)y» + σ(x)p(x)y = σ(x)q(x) The σ(x) we are goïg ë use makes ê left hå side ç this equation ê EXACT DERIVATIVEèç σ(x)yèi.e. [σ(x)y]» = σ(x)q(x) Integratïg both sides ç ê equation, usïg ê fact that ê left hå side is an exact derivative yields ░ σ(x)y =è▒èσ(x)q(x) dxè+èC ▓ Solvïg for y yields ê GENERAL SOLUTION ç ê lïear, first order differential equation èèè 1èíè░ è┐ 2) y =è──── ▒è▒ σ(x)q(x) dxè+èCè▒ èè σ(x) └è▓ è┘ The INTEGRATING FACTOR for this differential equation is èèèèèèèèèèèèèèèè èèèè èèèè ùèp(x) dxèèèèèèèèèèèèèèèè èèèè σ(x) =èe In computïg ê ïtegration facër, do NOT ïclude a constant ç ïtegration.èOnly one is necessary å it is ïcluded ï general solution 2). èè Formula 2) works ï all cases which means that this type ç differential equation always has a êoretical solution.èThere is a ê practical limitation that êre are two ïtegrals ïvolved.èIf eiêr or both cannot be ïtegrated ë an elementary function, ên êre will be no useful solution. èè The GENERAL SOLUTION will be valid for all x as long as ê functions p(x) å q(x) are contïuous everywhere.èPoïts ç disconïtuity will, ï general, cause ê solutions ë be ïvalid.èIn particular, ê ïtervals ç validity ç this solution will correspond ë ê ïterval ç contïuity ç ê functions p(x) å q(x). 1 y» - y = 3eì╣ A) 3 + Ce╣ B) 3e╣ + Ce╣ C) 3eì╣ + Ce╣ D) 3eÄ╣ + Ce╣ üèèèy» - y = 3eì╣ èèè èèThis differential equation is already ï lïear form, so ê ïtegratïg facër is èèèèèèèèèèèè ù (-1) dxèèèèèèèèèèèè σ(x) =èe =èeú╣ The second ïtegral becomes ░ ▒èeú╣ 3eì╣ dx ▓ ░ ▒è3e╣ dx ▓ This ïtegrates directly ë 3e╣ + C The general solution is èèè1 y =è───è[ 3e╣ + C ] èè eú╣ è=èe╣ [ 3e╣ + C ] y = 3eì╣ + Ce╣ ÇèC 2 xy» + y = xÅ A) xÅ/4 + C/x B) xÅ/4 + C ln[x] C) xÅ/5 + C/x D) xÅ/5 + C ln[x] üèèèxy» + y = xÅ èèè èèThis differential equation is NOT ï lïear form, so divide through by ê coefficient ç y» which is x, leavïg ê lïear differential equation èè 1 y» + ─ yè=èxÄ èè x so ê ïtegratïg facër is èèèèèèèèèèèè ù 1/x dx èèln[x]èèèèèèèèèèèèèèèèèèèèè σ(x) =èe =èeèèè =èx The second ïtegral becomes ░ ▒èx xÄ dx ▓ ░ ▒èxÅ dx ▓ This ïtegrates directly ë xÉ/5 + C The general solution is èèè1 y =è───è[ xÉ/5 + C ] èèèx è=èxÅ/5 + C/x ÇèC 3 èèè 2èèèè╨Ä y»è+è─ yè=èe èèè x ╨ÄèèèèèèèèèèèèèèxÄ A) eè /3x║è+èC/x║èèèB)èè eè /3xì + Cxì üèèèèèèè 2èèèxÄ èèèèèèèèy» + ─ y = e èèèèèèèèèè x èèè èèThis differential equation is already ï lïear form, so ê ïtegratïg facër is èèèèèèèèèèèè ù 2/x dxèèèèè 2 ln xèèèèèèèèèèèèèèèèèèèè σ(x) =èe =èe èèè èèèè ln[xì] èè =èeèèèèè èè =èxì The second ïtegral becomes ░èèè╨Ä ▒èxì eèèdx ▓ This ïtegral requires substitution u = xÄè du = 3xì dxèèdx = du/3xì ░èèèè du ▒èxì e╗ ─── ▓èèèè3xì 1è░ ─è▒èe╗ du 3è▓ This ïtegrates directly ë e╗ / 3 + C Substitutïg back ë ê origïal variable yields ╨Ä eè / 3 + C The general solution is èèè1è íè ╨Ä ┐ y =è───è▒èeè/3è+èC ▒ èèèxìè└èè ┘ èèè╨Ä è=èeè / 3xìè+èC / xì ÇèA 4 èèè 1 y»è+è─ yè= 3sï[x] èèè x A) 3cos[x] + 3sï[x]/x + C/x B) 3cos[x] - 3sï[x]/x + C/x C) -3cos[x] + 3sï[x]/x + C/x üèèèèèèè 1 èèèèèèèèy» + ─ y = 3sï[x] èèèèèèèèèè x èèè èèThis differential equation is already ï lïear form, so ê ïtegratïg facër is èèèèèèèèèèèè ù dx / xèèèèè ln[x]èèèèèèèèèèèèèèèèèèèèè σ(x) =èe =èeèèè èè =èx The second ïtegral becomes ░ ▒è3x sï[x]èdx ▓ This ïtegral requires ïtegration by parts u = 3xèèdu = 3dxèè dv = sï[x] dxèèv = -cos[x] The ïtegral becomes èèèèèèèèèèè ░èèèè - 3x cos[x] +è▒è3cos[x] dx èèèèèèè ▓èèè This ïtegrates directly ë - 3x cos[x] + 3sï[x] + C The general solution is èèè1è íè èèè ┐ y =è───è▒è- 3x cos[x] + 3sï[x] + C ▒ èèèxè └èè èèè ┘ èèè è=è- 3cos[x]è+ 3sï[x] / xè+èC / x ÇèC 5 y» + tan[x] yè=èsec[x] A) -sï[x] + C cos[x] B) sï[x] + C cos[x] C) cos[x] + C sï[x] D) -cos[x] + C sï[x] üèèèy» + tan[x] y = sec[x] èèThis differential equation is already ï lïear form, so ê ïtegratïg facër is èèèèèèèèèèèè ù tan[x] dxèèèèèèèèèèèè σ(x) =èe èèèèèèèèèèèè ù sï[x] dx / cos[x]èèèèèèèèèèèè èè =èeèèè This ïtegral requires substitution u = cos[x]è du = -sï[x] dxèè dx = - du / sï[x] èèèèèèèèèèèè ù - sï[x] du / {sï[x] u }èèèèèèèèèèèèè σ(x) =èe èèèèèèèèèèèèèè è- ù du / uèèèèèèèèèèèèèè èè =èe This ïtegrates directly ë è- ln[u] èè = e Substitutïg back ë ê origïal variable èln {cos[x]}úî èè = e èln {sec[x]} èè =èe èè = sec[x] The second ïtegral becomes ░ ▒èsec[x] sec[x]èdx ▓ èèèè░èèèè ▒èsecì[x] dx ▓èèè This ïtegrates directly ë tan[x] + C The general solution is èèèè1èèíè è ┐ y =è──────è▒ètan[x] + C ▒ èè sec[x]è└èè è ┘ èèè è=è tan[x]/sec[x]è+èC / sec[x] è=è sï[x] + C cos[x] ÇèB 6 y» + 2xyè=è2x èèè╨ì èèè A) 2 + Ce èèè-╨ì èèè B) 1 + Ce üèèèy» + 2xy = 2x èèè èèThis differential equation is already ï lïear form, so ê ïtegratïg facër is èèèèèèèèèèèè ù 2x dxèèèèèèèèèèèè σ(x) =èe ╨ì èè =èeèèè The second ïtegral becomes ░èèè╨ì ▒è2x eèèdx ▓ This requires substitution with u = xìè du = 2x dx èèèè░èèèè ▒èe╗ du ▓èèè This ïtegrates directly ë e╗ + C Substitutïg back ë ê origïal variable yields ╨ì eè + c The general solution is èèèèè╨ì íè ╨ì ┐ y =è1 / eè ▒èeè + C ▒ èèèèèè └èè ┘ èèè èè -╨║ è=è1 +èC e ÇèB äèèSolve ê ïitial value problem â For ê lïear, first order differential equation y' + y = xèèy(0) = 5 ê ïtegratïg facër isèσ(x) = e╣.è The general solution isèy = x - xeú╣ + Ceú╣ Substitutïg x = 0 ïë ê general solution yields 5 = C so ê ïitial value problem's solution isèèy = x - xeú╣ + 5eú╣ éS èèA full discussion ç Initial Value Problems for FIRST ORDER DIFFERENTIAL EQUATIONS is ï Section 1.2.è èèBriefly, solvïg an Initial Value Problem is a two-step process.èFirst, fïd ê GENERAL SOLUTION ç ê differential equation.è Second, substitute ï ê ïitial value ïfor- mationèi.e.èx╠ for x å y╠ for y.èThis will produce an equation for C which provides ê value ç ê arbitrary constant ë put back ï ê general solution. 7 xy» + y =èxe╣ y(1) = 5e A) e╣ + e╣/x + 5e/x B) e╣ - e╣/x + 5e/x C) eú╣ + eú╣/x + 5e/x D) eú╣ - eú╣/x + 5e/x üèèèxy» + y = xe╣, y(1) = 5e èè èèThis differential equation is not ï lïear form, so divide ê entire equation by ê coefficient ç y» i.e. by x ë yield èè 1 y» + ─ yè=èe╣ èè x The ïtegratïg facër is èèèèèèèèèèèè ùè1 / x dxèèèèèèèèèèèè σ(x) =èe ln[x] èè =èeèèè èè èè =èx The second ïtegral becomes ░èèè ▒èx e╣èdx ▓ This requires ïtegration by parts with u = xèèèèdu = dx dv = e╣dxèèv = e╣ èèèèèèèè░èèèè xe╣è-è▒èe╣ dx èèèè ▓èèè This ïtegrates directly ë xe╣ - e╣ + C The general solution is èè 1èíèèèèèèèè┐ y =è─è▒èxe╣ - e╣è+ C ▒ èè xè└èè èèèè ┘ èèè èèè è=èe╣è-èe╣ / xè+èC / x As ê ïitial condition is y(1) = 5e, substitute x = 1 å èèèèy = 5e. 5e = eîè- eî/1 + C/1 Solvïg yieldsè C = 5e Thus ê specific solution is y = e╣ - e╣/x + 5e/x ÇèB 8 y» - 2yè=è2eÅ╣ y(0) = 4 A) eÅ╣ + 3eì╣ B) 2eÅ╣ + 2eì╣ C) -eÅ╣ + 3eì╣ D) 2eÅ╣ + 2 üèèèy» - 2y = 2eÅ╣, y(0) = 4 èèè èèThis differential equation is already ï lïear form, The ïtegratïg facër is èèèèèèèèèèèè ùè-2 dxèèèèèèèèèèèè σ(x) =èe èè =èeúì╣èèè The second ïtegral becomes ░èèè ▒è2eÅ╣ eúì╣èdx ▓ èèèè░èèèè ▒ 2eì╣ dx èèèè▓èè This ïtegrates via substitution usïg u = 2xèè du = 2 dxèèdx = du/2 ░ ▒ 2 e╗èdu/2 ▓ ░ │èe╗ du ▓ This ïtegrates directly ë e╗ + C Substitutïg back ë ê origïal variable yields eì╣ + C The general solution èèè1 y = ────è[èeì╣ + C ▒ èèeúì╣èèèè èèè è=èeÅ╣è+èCeì╣ As ê ïitial condition is y(0) = 4, substitute x = 0 å èèèèy = 4. 4 = eòè+èCeò Solvïg yieldsè C = 3 Thus ê specific solution is y = eÅ╣ + 3eì╣ ÇèA 9 y» + cot[x]yè=èxì y(π/2) = 3π A) -xìcot[x] + 2x + 2cot[x] + 2πcsc[x] B) xìcot[x] - 2x + 2cot[x] + 2πcsc[x] C) xìcot[x] + 2x - 2cot[x] + 2πcsc[x] D) xìcot[x] + 2x + 2cot[x] - 2πcsc[x] üèèèy» + cot[x]y = xì, y(π/2) = 3π èèThis differential equation is already ï lïear form, The ïtegratïg facër is èèèèèèèèèèèè ùècot[x] dxèèèèèèèèèèèè σ(x) =èe ù cos[x] dx / sï[x]èèèèèèèèèèèèè èè =èe This requires substitution with u = sï[x]è du = cos[x dx this yields èèèèèèèèèèèèè ùèdu / uèèèèèèèèèèèè èè =èe This ïtegrates directly ë ln{sï[x]} èè =èeèè èè =èsï[x] The second ïtegral becomes ░èèè ▒èsï[x] xì dx ▓ This requires ïtegration by parts u = xì èèdu = 2x dx dv = sï[x]èv = -cos[x] èèèèèèèèèèèèè░èèèè =è- xì cos[x]è+ ▒ 2x cos[x] dx èèèèèèèèèèèèè▓èè This ïtegral also requires ïtegration by parts u = 2xèèèdu = 2 dx dv = cos[x]èv = sï[x] èèè ░ =è- xì cos[x]è+ 2x sï[x] -è▒ 2 sï[x] dx èèè ▓ This ïtegrates directly ë =è- xì cos[x]è+è2x sï[x]è+è2 cos[x]è+èC The general solution is èèè 1 y = ──────è[ -xìcos[x] + 2xsï[x] + 2cos[x] + C ] èèsï[x] èèèè èèè è=è-xì cot[x] + 2x + 2cot[x] + Ccsc[x] As ê ïitial condition is y(π/2) = 3π, substitute x = π/2 èèèèå y = 3π. 3π = -πì/4 cot[π/2] + 2(π/2) + 2 cot[π/2] + Ccsc[π/2] 3π = π + C Solvïg yieldsè C = 2π Thus ê specific solution is y = -xì cot[x] + 2x + 2cot[x] + 2π csc[x] ÇèA