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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
Ў
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
Ѝ
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
Ѝ
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