first-order di erential equationsfacstaff.cbu.edu/wschrein/media/m231 notes/m231c2.pdf18 2....

CHAPTER 2

First-Order Di↵erential Equations

1. Introduction: Motion of a Falling Body

Problem. An object falls through the air toward earth. Assuming that theonly forces acting on the object are gravity and air resistance, determine thevelocity of the object as a function of time.

With F the total force on the object, m the mass and v the velocity of the

object,dv

dtgives the acceleartion of the object. By Newton’s second law,

mdv

dt= F.

Here we will assume v is positive when it is directed downward. Also, near theEarth’s surface, the force due to gravity is mg where g is the acceleration due togravity. Air resistance, which is proportional to velocity, is given by �bv whereb is a positive constant depending on the density of the air and the shape ofthe object. The negative sign is since the air resistance acts opposite to gravity.Thus we have the first order DE

mdv

dt= mg � bv.

We solve this equation using separation of variables, treating dv and dt as dif-ferentials. Assuming m 6= 0 and mg � bv 6= 0, we can get

dv

mg � bv=

dt

m

17

18 2. FIRST-ORDER DIFFERENTIAL EQUATIONS

Integrating, Zdv

mg � bv=

Zdt

m+ c.

We always add the constant at the integration step. NotingZ

dv

mg � bv= �1

b

Z �b

mg � bvdv,

so the numerator of the integrand is the derivative of the denominator, we get

� 1

bln|mg � bv| =

t

m+ c =) ln|mg � bv| = �bt

m� bc =)

|mg � bv| = e�btme�bc =) mg � bv = ±e�bce�

btm =)

mg � bv = Ae�btm (A 6= 0).

Solving for v,

(⇤) v =mg

b� A

be�

btm ,

a general solution to the DE. For a specific solution, we solve the IVP

mdv

dt= mg � bv, v(0) = v0,

1. INTRODUCTION: MOTION OF A FALLING BODY 19

where v0 is the initial velocity. From (⇤),

A = �v0b + mg =) v =mg

b+

✓v0 �

mg

b

◆e�

btm .

Since b > 0, limt!1

e�btm = 0 =) lim

t!1v =

mg

b.

The constantmg

bis referred to as the limiting or terminal velocity of the object.

This can be seen in the graph below.

The simplest 1st-order equations are of the form

dy

dx= f(x).

For these,

y(x) =

Zf(x) dx + C

are solutions.

We shall expect all students know the Standard Integral Forms and Formulason the handout provided.


Example.

(1)dy

dx=

5x + 6

x2 + 4

y =

Z5x + 6

x2 + 4dx =

Z5x

x2 + 4dx +

Z6

x2 + 4dx =

5

2

Z2x

x2 + 4dx +

Z12

4u2 + 4du =

x=2u

dx=2du

5

2

Z2x

x2 + 4dx + 3

Z1

u2 + 1du =

5

2ln(x2 + 4) + 3 arctanu + C =

5

2ln(x2 + 4) + 3 arctan

⇣x

2

⌘+ C

(2)dy

dx=

1p1� 4x2

y =

Z1p

1� 4x2dx =

1

2

Z1p

1� u2du =

u=2x

du=2dx

1

2arcsinu + C =

1

2arcsin(2x) + C


(3)dy

dx= xe5x

y =

Zxe5x dx =

Zu dv = uv �

Zv du (integration by parts)

u = dv =

du = v =

u = x dv = e5x dx

du = dx v = 15e

5x

x

5e5x � 1

5

Ze5x dx =

x

5e5x � 1

25e5x + C =

e5x

25(5x� 1) + C

In choosing which function to use for u, one can use LIATE:

L = log

I = inverse trig

A = algebraic

T = trig

E = exponential


(4)dx

dt= e�3t sin 5t

x =

Ze�3t sin 5t dt =

u = sin 5t dv = e�3t dt

du = 5 cos 5t dt v = �13e�3t

�1

3e�3t sin 5t +

5

3

Ze�3t cos 5t dt =

u = cos 5t dv = e�3t dt

du = �5 sin 5t dt v = �13e�3t

�1

3e�3t sin 5t +

5

3

h� 1

3e�3t cos 5t� 5

3

Ze�3t sin 5t dt

i=)

34

9

Ze�3t sin 5t dt = �1

3e�3t

hsin 5t +

5

3cos 5t

i=)

Ze�3t sin 5t dt = � 3

34e�3t

hsin 5t +

5

3cos 5t

i+ C


Tabular approach to integration by parts:

(4)

x =

Ze�3t sin 5t dt =

sin 5t e�3t

+&

5 cos 5t �13e�3t

�&

�25 sin 5t+ 1

9e�3t

�1

3e�3t sin 5t� 5

9e�3t cos 5t� 25

9

Ze�3t sin 5t dt

Then continue as above.

(3)

y =

Zxe5x dx =

x e5x

+&

1 15e

5x

�&

0+ 1

25e5x

1

5xe5x � 1

25e5x +

Z0 dx

Again, continue as above.


(5)dy

dx=

2x3 � 2x2 � 65x� 107

x2 � 5x� 14

y =

Z2x3 � 2x2 � 65x� 107

x2 � 5x� 14dx =

Z(2x + 8) dx +

Z3x + 5

(x� 7)(x + 2)dx =

————————————————————3x + 5

(x� 7)(x + 2)=

A

x� 7+

B

x + 2

3x + 5 = A(x + 2) + B(x� 7)

A + B = 3 A =26

92A� 7B = 5 B =

1

97A + 7B = 21——————–9A=26

————————————————————

x2 + 8x +26

9

Zdx

x� 7+

1

9

Zdx

x + 2=

x2 + 8x +26

9ln|x� 7| +

1

9ln|x + 2| + C


(6)dy

dx=

x3 � 5x2 + 10x� 9

(x� 2)2(x2 � 3x + 1)

y =

Zx3 � 5x2 + 10x� 9

(x� 2)2(x2 � 3x + 1)dx =

————————————————————x3 � 5x2 + 10x� 9

(x� 2)2(x2 � 3x + 1)=

A

x� 2+

B

(x� 2)2+

Cx + D

x2 � 3x + 1

x3 � 5x2 + 10x� 9 =

A(x� 2)(x2 � 3x + 1) + B(x2 � 3x + 1) + (Cx + D)(x� 2)2 =

Ax3�5Ax2+7Ax�2A+Bx2�3Bx+B+Cx3�4Cx2+Dx2+4Cx�4Dx+4D

A + C = 1 A = �1�5A + B � 4C + D = �5 B = 17A� 3B + 4C � 4D = 10 C = 2�2A + B + 4D = �9 D = �3

What about x2 in the denominator? Can use eitherAx + B

x2or

A

x+

B

x2.

————————————————————

�Z

dx

x� 2+

Z1

(x� 2)2dx +

Z2x� 3

x2 � 3x + 1dx =

� ln|x� 2|

u=x-2

du=dx

+

Zu�2 du + ln

��x2 � 3x + 1�� =

ln��x2 � 3x + 1

�� ln|x� 2|� 1

x� 2+ C = ln

��x2 � 3x + 1

x� 2

��1

x� 2+ C


(7)dy

dx=

5

x2 + 6x + 10

y =

Z5

x2 + 6x + 10dx =

Z5

(x2 + 6x + 9) + 1dx = 5

Z1

(x + 3)2 + 1dx =

u=x+3

du=dx

5

Z1

u2 + 1du =

5 arctanu + C = 5 arctan(x + 3) + C

(8) The familiar (I hope) identities

1� sin2 ✓ = cos2 ✓, 1 + tan2 ✓ = sec2 ✓, sec2 ✓ � 1 = tan2 ✓

are used in simplifying algebraic expressions in integrals by trigonometric

substitution. The three substitutions used:

(I)p

a2 � x2, set x = a sin ✓(II)

pa2 + x2, set x = a tan ✓

(III)p

x2 � a2, set x = a sec ✓


dy

dx=

x2

px2 � 16

y =

Zx2

px2 � 16

dx =

√(x2 -16)x

4 x = 4 sec ✓, dx = 4 sec ✓ tan ✓d✓p

x2 � 16 =p

16 sec2 ✓ � 16 = 4p

sec2 ✓ � 1 = 4 tan ✓Z

16 sec2 ✓

4 tan ✓(4 sec ✓ tan ✓)d✓ = 16

Zsec3 ✓ d✓ =

sec ✓ sec2 ✓+&

sec ✓ tan ✓� tan ✓

16

✓sec ✓ tan ✓ �

Zsec ✓ tan2 ✓ d✓

◆=

16


Zsec ✓(sec2 ✓ � 1) d✓

◆=

16


Z(sec3 ✓ � sec ✓) d✓

◆=

16 sec ✓ tan ✓ � 16

Zsec3 ✓ d✓ + 16

Zsec ✓ d✓ =)

32

Zsec3 ✓ d✓ = 16 sec ✓ tan ✓ + 16 ln|sec ✓ + tan ✓| =)


y = 8 sec ✓ tan ✓ + 8 ln|sec ✓ + tan ✓| =)

√(x2 -16)x

4

y = 8 · x

4·p

x2 � 16

4+ 8 ln

��x

4+

px2 � 16

4

�� =)

y =xp

x2 � 16

2+ 8 ln

��x

4+

px2 � 16

4

�� + C

Maple. See integration.mw or integration.pdf.

2. SEPARABLE EQUATIONS 29

2. Separable Equations

Definition (1 — Separable Equation). A first-order di↵erential equation

dy

dx= f(x, y)

is said to be separable if

f(x, y) = g(x) · p(y),

where g depends only on x and p depends only on y.Example.

(1)dy

dx=

x� 5

y2= (x� 5) · 1

y2, so this equation is separable. To solve,

y2 dy = (x� 5) dx =)Z

y2 dy =

Z(x� 5) dx + K =)

y3

3=

x2

2� 5x + K

| {z }implicit solution

=) y3 =3

2x2 � 15x + 3K =)

y =h3

2x2 � 15x + 3K

i1/3=) y =

h3

2x2 � 15x + C

i1/3

| {z }explicit solution

.

You can graph this solution on a calculator for various values of C. To checkthat y is the solution,

dy

dx=

1

3

h3

2x2 � 15x + C

i�2/3(3x� 15)

=x� 5

h⇣32x

2 � 15x + 3K⌘1/3i2

=x� 5

y2.

(2)dy

dx= 1 + xy

is not separable.


(3)dy

dx=

y � 1

x + 3=

1

x + 3· (y � 1)

(a) y � 1 = 0 =) dy

dx= 0 =)

y = 1 is a constant solution.

(b) For y 6= 1, we can divide by y � 1, so

dy

y � 1=

dx

x + 3=)

Zdy

y � 1=

Zdx

x + 3+ K =)

ln|y � 1| = ln|x + 3| + K (an implicit solution) =)eln|y�1| = eln|x+3|+K = eln|x+3|eK =)

|y � 1| = |x + 3| · K1 (K1 > 0) =)y � 1 = ±K1(x + 3) =)

y � 1 = C(x + 3) (C 6= 0) =)y = 1 + C(x + 3).

We can put (a) and (b) together here to get

y = 1 + C(x + 3) (C 2 R)

as a general solution. This latter step may not always be able to be done.


(4)dy

dx=

6x5 � 2x + 1

cos y + ey= (6x5 � 2x + 1) · 1

cos y + ey. We have

(cos y + ey) dy = (6x5 � 2x + 1) dx.

Then Z(cos y + ey) dy =

Z(6x5 � 2x + 1) dx + K =)

sin y + ey = x6 � x2 + x + K| {z }implicit solution

We check using implicit di↵erentiation:

(cos y)y0 + eyy0 = 6x5 � 2x + 1 =)y0(cos y + ey) = 6x5 � 2x + 1 =)

dy

dx= y0 =

6x5 � 2x + 1

cos y + ey

Indefinite Integral Convention for Di↵erential Equations

The indefinite integral

Zg(x) dx represernts any (single) antiderivative of g(x).

In other words, it means any function G(x) with the derivative

d

dxG(x) = g(x).


Why Separation of Variables works

Assume g(x) and h(y) are continuous functions, that

dy

dx= g(x) · h(y),

and that h(y) 6= 0 on the interval where the solution is sought. Then

1

h(y)

dy

dx= g(x).

Let

r(y) =1

h(y)=)

(#) r(y)dy

dx= g(x).

Let

R(y) =

Zr(y) dy =) dR

dy= r(y)

and

G(x) =

Zg(x) dx =) dG

dx= g(x).

Then, from (#),

dR

dy· dy

dx=

dG

dx=) (from the chain rule)

d

dx

hR

�y(x)

�i=

d

dxG(x) =)

R(y) = G(x) + C =)Z

r(y) dy =

Zg(x) dx + C,

an implicit solution. If possible, we then get an explicit solution by solving fory in terms of x.


Example (An initial value problem (IVP)). In an IVP, besides the di↵er-ential equation, we are given an initial condition that allows us to find a specificsolution rather than a general solution with a constant of integration.

dy

dx= 8x3e�2y, y(0) = 0

Because of the initial condition, the only possible constant solution is y = 0,but that is not a solution here. If it were, we would be done since there is onlya single solution to an IVP. We have

e2y dy = 8x3 dx =)Z

e2y dy =

Z8x3 dx + K =)

1

2e2y = 2x4 + K =) e2y = 4x4 + 2K =)

ln e2y = ln(4x4 + 2K) =) 2y = ln(4x4 + C) =)

y =1

2ln(4x4 + C)

This is the general solution. Then

y(0) =1

2ln C = 0 =)

ln C = 0 =)C = 1

Thus the solution to the IVP is

y =1

2ln(4x4 + 1) = ln

p4x4 + 1


Example (Using definite integration to solve an IVP).

dy

dx= 8x3e�2y, y(0) = 0

As before, we havee2y dy = 8x3 dx.

We use Z y

0e2r dr =

Z x

08s3 ds.

Since x and y become the upper limits of integration here, we use r and s asdummy variables for the integrals. The initial condition says

x = 0 =) y = 0.

The value of x from the initial condition becomes the lower limit of the “x”integral and the value of y from the initial condition becomes the lower limit ofthe “y” integral. Continuing,

1

2e2r

��r=y

r=0= 2s4

��s=x

s=0=)

1

2e2y � 1

2= 2x4 =)

e2y = 4x4 + 1 =)2y = ln(4x4 + 1) =)

y =1

2ln(4x4 + 1) = ln

p4x4 + 1

Maple. See separable.mw or separable.pdf


Example. A raindrop is falling through the atmosphere. Aside from theforce of gravity, assume all other forces acting on the raindrop are negligible.Suppose that the atmosphere is saturated with water vapor and that as a resultof condensation, the mass of the raindrop is increasing at a rate proportional toits surface area. Assume the raindrop is always spherical and that its density⇢ remains constant. Let its radius at time t = 0 be r0 and its velocity v0.

(a) Show that the radius of the drop increases linearly with time.

Let V = V (t) = the volume at time t, m = m(t) = the mass at time t.

[To find an equation involving r and t.]

We know

m = V ⇢ =4

3⇡r3⇢.

We are givendm

dt/ SA = 4⇡r2 =)

dm

dt= �(4⇡r2), � > 0.

Then substituting for m,

d

dt

h4

3⇡r3⇢

i= 4�⇡r2 =) (chain rule)

4⇡r2⇢dr

dt= 4�⇡r2 =)

dr

dt=

�

⇢.

Letting k =�

⇢> 0,

dr

dt= k.

Thenr = kt + r0.


(b) Express the velocity v = v(t) of the raindrop in terms of its radius r.

Let p = p(t) = the momentum of the drop.

By Newton’s 2nd law of motion,

dp

dt= Fsum

where p = mv and Fsum = ma = �mg. Then

dp

dt=

d

dt(mv) = �mg =)

d

dt

h4

3⇡r3⇢v

i= �4

3⇡r3⇢g =) 4

3⇡⇢

d

dt(r3v) = �4

3⇡r3⇢g =)

d

dt(r3v) = �r3g.

Now, since r is increasing, r has an inverse, i.e., t is a function of r, so v = v(t)is also a function of r. Then, from the chain rule,

d

dr(r3v)

dr

dt= �r3g =) (from previous page)

kd

dr(r3v) = �r3g =) d

dr(r3v) = �g

kr3 =)

Zd

dr(r3v) dr =

Z ⇣� g

kr3

⌘dr + C =)

r3v = �g

k· r4

4+ C.

At t = 0 (with v0 = v(0)),

r30v0 = �gr4

0

4k+ C =) C = r3

0v0 +gr4

0

4k.

Then

v = � g

4kr +

1

r3

⇣r30v0 +

gr40

4k

⌘=) v = � g

4kr +

⇣r0

r

⌘3⇣v0 +

gr0

4k

⌘.

3. LINEAR EQUATIONS 37

3. Linear Equations

Definition. A first-order DE is linear if it can be expressed in the form

a1(x)dy

dx+ a0(x)y = b(x)

where a1(x), a0(x), and b(x) are arbitrary functions of x.Example.

(1) x2 sin x + (cos x)y = sin xdy

dx=)

sin xdy

dx� (cos x)y = x2 sin x, and this is linear.

(2) ydy

dx+ (sin x)y3 = ex + 1

is not linear

Two easy cases

(1) a0(x) = 0 =) a1(x)dy

dx= b(x) =)

dy

dx=

b(x)

a1(x), a1(x) 6= 0 =)

y =

Zb(x)

a1(x)dx + C.


(2) a01(x) = a0(x). Then

a1(x)dy

dx+ a01(x)y = b(x) =)

d

dx

ha1(x)y

i= b(x) =)

a1(x)y =

Zb(x) dx + C =)

y =1

a1(x)

h Zb(x) dx + C

i.

Suppose a1(x) 6= 0 for all x on some interval. Then

dy

dx+

a0(x)

a1(x)y =

b(x)

a1(x)or

(⇤) dy

dx+ P (x)y = Q(x),

the standard form of a first-order linear DE.

Is (⇤) separable?

(1) if either P (x) ⌘ 0 or Q(x) ⌘ 0.

(2) if both P (x) and Q(x) are constants.

(3) if P (x) and Q(x) are constant multiples of each other.


Simplest case: P (x) ⌘ r (a constant), Q(x) = 0.

y0 + ry = 0

Multiply through the equation by erx. We call erx an integrating factor. Notethat this factor will never be 0, so no extraneous solutions will be introduced.

erxy0 + rerxy = 0 =)d

dx

herxy

i= 0 =)

erxy = C (we integrated here) =)y(x) = Ce�rx

We check our solution:d

dx

⇣Ce�rx

⌘+ rCe�rx = �rCe�rx + rCe�rx = 0

Next case: y0 + ry = Q(x), Q(x) continuous.

We again multiply through by the integrating factor erx.

erxy0 + rerxy = erxQ(x) =)d

dx

herxy

i= erxQ(x) =)

erxy =

ZerxQ(x) dx + C =)

y(x) =1

erx

"ZerxQ(x) dx + C

#


Example.

(1)dy

dx� y = e3x.

Since r = �1, the integrating factor is e�x:

e�xy0 � e�xy = e�xe3x =)d

dx

he�xy

i= e2x =)

e�xy =

Ze2x dx + C =)

e�xy =1

2e2x + C =)

y(x) =1

2e3x + Cex

(2)dx

dt+ 4x = e�t, x(0) =

4

3.

Since r = 4, the integrating factor is e4t:

e4tx0 + 4e4tx = e4te�t =)d

dt

he4tx

i= e3t =)

e4tx =

Ze3t dt + C =)

e4tx =1

3e3t + C =)

x(t) =1

3e�t + Ce�4t.

x(0) =1

3+ C =

4

3=) C = 1 =)

x(t) =1

3e�t + e�4t =

e�4t

3

⇣e3t + 3

⌘=

e3t + 3

3e4t


General case:dy

dx+ P (x)y = Q(x), Q(x) continuous.

We need to find an integrating factor µ(x). We multiply the equation by µ(x).

(⇤) µ(x)dy

dx+ µ(x)P (x)| {z } y = µ(x)Q(x)

We want the left-hand side of (⇤) to to be the derivative of some product, i.e.,we want

d

dxµ(x) = µ(x)P (x)

or

(#)dµ

dx= µP (x).

But this separable, so

dµ

µ= P (x) dx =)

Zdµ

µ=

ZP (x) dx + K =)

ln|µ| =

ZP (x) dx + K =)

|µ| = eR

P (x) dx+K = eR

p(x) dxeK =)µ = Ce

RP (x) dx (C 6= 0)

Let µ = eR

P (x) dx be the integrating factor.

This works because it is a solution of (#). (⇤) then becomes

(⇤) d

dx

hµ(x)y

i= µ(x)Q(x) =)

µ(x)y =

Zµ(x)Q(x) dx + C =)

y(x) =1

µ(x)

"Zµ(x)Q(x) dx + C

#


Now suppose P (x) and Q(x) are continuous on (a, b) and a < x0 < b. Also,suppose y(x0) = y0.

Let

W (x) =

ZP (x) dx (a family of functions with W 0(x) = P (x))

and

F (x) =

Z x

x0

P (s) ds = W (x)�W (x0)

for a < x < b. Then

F 0(x) = W 0(x) = P (x) and F (x0) = 0.

So choose ZP (x) dx =

Z x

x0

P (s) ds.

Thenµ(x) = e

R xx0

P (s) ds and µ(x0) = e0 = 1.

Then, from (⇤),Z x

x0

d

ds

hµ(s)y(s)

ids =

Z x

x0

µ(s)Q(s) ds =)

µ(x)y(x)� µ(x0)y(x0) =

Z x

x0

µ(s)Q(s) ds =)

y(x) =1

µ(x)

"Z x

x0

µ(s)Q(s) ds + y0

#

is the solution.

We have shown:


Theorem (1 — Existence and Uniqueness of Solution). Let P (x) and Q(x)be continuous functions on an interval (a, b) and let x0 be any point in thisinterval. Then the initial value problem

dy

dx+ P (x)y = Q(x), y(x0) = y0

has a unique solution on the interval (a, b).Example.

1

x

dy

dx� 2y

x2= x cos x, x > 0. We then have

y0 � 2

xy = x2 cos x.

µ = eR

(� 2x) dx = e�2

R dxx = e�2 lnx = elnx�2

= x�2 =1

x2

Then1

x2y0 � 2

x3y = cos x =)

d

dx

hx�2y

i= cos x =)

x�2y =

Zcos x dx + C =)

x�2y = sin x + C =)y(x) = x2(sin x + C)

We check:1

x

dy

dx� 2y

x2=

1

x

⇥2x(sin x + C) + x2 cos x

⇤� 2(sin x + C) =

2(sin x + C) + x cos x� 2(sin x + C) = x cos x


Problem (Page 52 # 24a). A rock contains two radioactive isotopes, RA1

and RA2, that belong to the same radioactive series; that is, RA1 decays intoRA2, which then decays into stable atoms. Assume that the rate at whichRA1 decays into RA2 is 50e�10t kg/sec. Because the rate of decay of RA2 isproportional to the mass y(t) of RA2 present, the rate of change in RA2 is

dy

dt= rate of creation� rate of decay,

dy

dt= 50e�10t � ky,

where k > 0 is the decays constant. If k = 20/sec and initially y(0) = 40 kg,find the mass y(t) of RA2 for t > 0. The IVP in standard form is then

dy

dt+ 20y = 50e�10t, y(0) = 40.

µ = eR

20 dt = e20t.

Then

e20ty0 + 20e20ty = 50e10t =)d

dt

he20ty

⇤= 50e10t =)

e20ty = 50

Ze10t dt + C =)

e20ty = 50⇣ 1

10

⌘e10t + C =)

y(t) = 5e�10t + Ce�20t.

y(0) = 5 + C = 40 =) C = 35.

Thusy(t) = 5e�10t + 35e�20t.


Example. (1 + t2)x0 + 4tx = (1 + t2)�2, x(0) = 5. Then

x0 +4t

1 + t2x = (1 + t2)�3.

µ = eR 4t

1+t2dt

= e2R 2t

1+t2dt

= e2 ln(1+t2) = eln(1+t2)2 = (1 + t2)2

(1 + t2)2x0 + 4t(1 + t2)x =1

1 + t2=)

d

dt

h(1 + t2)2x

i=

1

1 + t2=)

(1 + t2)2x =

Z1

1 + t2dt + C =)

(1 + t2)2x = arctan t + C =)

x(t) =arctan t + C

(1 + t2)2.

x(0) =C

1= 5 =) C = 5.

Thus

x(t) =arctan t + 5

(1 + t2)2.


Problem (Page 142 #33).dx

dt+ 2tx = 1, x(0) = 1

µ = eR

2t dt = et2

et2x0 + 2tet2x = et2 =)d

dt

het2x

i= et2 =) et2x =

Zet2 dt + C.

There is no elementary formula for

Zet2 dt, so we instead integrate each side

from t0 = 0 to t.Z t

0

d

ds

hes2

xids =

Z t

0es2

ds =)

es2x(s)

��t

0=

Z t

0es2

ds =)

et2x(t)� e0 · 1 =

Z t

0es2

ds =)

x(t) = e�t2h Z t

0es2

ds + 1i

=)

x(t) = e�t2Z t

0es2

ds + e�t2 =)

Dawson(t) = e�t2Z t

0es2

ds and erf(t) =2

⇡

Z t

0es2

ds

x(t) = Dawson(t) + e�t2.

Maple. See first order linear.mw or first order linear.pdf

4. EXACT EQUATIONS 47

4. Exact Equations

Maple. See function 2 variable.mw or function 2 variable.pdf

Partial Derivatives

First-order partial derivatives: Consider a function F (x, y) defined on a regionR 2 R2. Let (a, b) be an interior point of R. The average rate of change asyou move horizontally from (a, b) to (a + h, b) is

F (a + h, b)� F (a, b)

h.

The instantaneous rate of change in the x-direction at (a, b) is

@F

@x(a, b) = lim

h!0

F (a + h, b)� F (a, b)

h,

the partial derivative of F with respect to x.


The average rate of change as you move vertically from (a, b) to (a, b + h) is

F (a, b + h)� F (a, b)

h.

The instantaneous rate of change in the y-direction at (a, b) is

@F

@y(a, b) = lim

h!0

F (a, b + h)� F (a, b)

h,

the partial derivative of F with respect to y.

Example. Let F (x, y) = x2y2.

@F

@x(x, y) = lim

h!0

F (x + h, y)� F (x, y)

h= lim

h!0

(x + h)2y2 � x2y2

h

= limh!0

x2y2 + 2xhy2 + h2y2 � x2y2

h= lim

h!0

2xhy2 + h2y2

h= lim

h!0(2xy2 + hy2) = 2xy2.

Basically, hold y constant and take the derivative with respect to x. We do

similarly for@F

@y(x, y). Then, for example,

@F

@x(3, 2) = 2 · 3 · 22 = 24.


Notation.@F

@x(x, y)

| {z }traditional notation

= Fx(x, y)| {z }modern notation

=@

@x

⇥F (x, y)

⇤

| {z }partial di↵erential operator

z }| {@F

@y(x, y) =

z }| {Fy(x, y) =

z }| {@

@y

⇥F (x, y)

⇤

Example.

@

@x(xep

xy) =@

@x(xe(xy)1/2) = e(xy)1/2 + xe(xy)1/2

⇣1

2

⌘(xy)�1/2(y) =

⇣1

2

⌘ep

xy(2 +p

xy).

Note.x �!

yxykz

�!12z�1/2

(xy)1/2

k

z1/2

ks

�!es

e(xy)1/2

k

ez1/2

kes

Example. F (x, y) = x ln(y cos x). Find@F

@x

⇣⇡

3, 1

⌘.

@F

@x(x, y) = ln(y cos x) + x

1

y cos x(�y sin x) = ln(y cos x)� x tan x

Note.x �!� sin x

cos xks

�!y

y cos xkyskz

�!1z

ln(y cos x)k

ln(ys)k

ln z

Thus@F

@x

⇣⇡

3, 1

⌘= ln

1

2� ⇡

3

p3 = �

⇣ln 2 +

⇡p3

⌘.


2nd Order Partial Derivatives

Fxx =@

@xFx =

@

@x

@F

@x=

@2F

@x2

Fxy =@

@yFx =

@

@y

@F

@x=

@2F

@y@x

Fyx =@

@xFy =

@

@x

@F

@y=

@2F

@x@y

Fyy =@

@yFy =

@

@y

@F

@y=

@2F

@y2

Note. This method also extends to more variables and higher derivatives.

Example. F (x, y) = 3x4y5 + 8x2 � 5y + 15

Fx(x, y) = 12x3y5 + 16x

Fy(x, y) = 15x4y4 � 5

Fxx(x, y) = 36x2y5 + 16

Fxy(x, y) = 60x3y4

Fyx(x, y) = 60x3y4

Fyy(x, y) = 60x4y3

Note. Notice that Fxy = Fyx.

Theorem (8.1). If Fxy and Fyx are continuous at an interior point (a, b)of their domain, then

Fxy(a, b) = Fyx(a, b).

Maple. See partderiv.mw or partderiv.pdf


Exact Di↵erential Equations

Suppose we coordinatize a map of the US so each point on the map has an (x, y)coordinate pair. Then consider the Fahrenheit temperature function F (x, y)defined at each point of the map for a given time on a given day.

We see that temperature changes continuously and smoothly across the US.These curves F (x, y) = C, where C = 50, 60, 70, 80, 90 on the graph above,are called level curves of the graph. For weather maps, these level curves arecalled isotherms. Because the level curves are smooth, having tangent lines(derivatives) at each point, they are actually implicit solutions of di↵erentialequations.

We find the slope of the tangent to a level curve by implicit di↵erentiation.Assuming y is a function of x, we take the derivative of F (x, y) = C implicitly:

d

dxF (x, y) =

d

dx(C) =)

(⇤) @F

@x+

@F

@y

dy

dx= 0 =)

(⇤⇤) dy

dx= �@F/@x

@F/@y, the slope at (x, y).


Multiplying the left-hand side of (⇤) by dx yields the total di↵erential dF :

dF :=@F

@xdx +

@F

@ydy or dF := Fx dx + Fy dy.

Setting dF = 0 and solving allows us to obtain the equation for the slopef(x, y) of the level curve F (x, y) = C. Since (⇤⇤) is a di↵erential equation, weshould be able to reverse the logic and easily solve some DE’s. Note that any

first-order DEdy

dx= f(x, y) can be rewritten in the (di↵erential) form

(#) M(x, y) dx + N(x, y) dy = 0.

If the left-hand side is a total di↵erential

M(x, y) dx + N(x, y) dy = Fx dx + Fy dy = dF (x, y),

then its solutions are given (implicitly) by F (x, y) = C, where C is an arbitraryconstant.

Definition (2 — Exact Di↵erential Form).

The di↵erential form M(x, y) dx+N(x, y) dy is said to be exact in a rectangleR if there is a function F (x, y) such that

@F

@x(x, y) = M(x, y) and

@F

@y(x, y) = N(x, y)

for all (x, y) 2 R. That is, the total di↵erential of F (x, y) satifies

dF = M(x, y) dx + N(x, y) dy.

If M(x, y) dx + N(x, y) dy is an exact di↵erential form, then the equation

M(x, y) dx + N(x, y) dy = 0

is called an exact equation.


How can we judge whether a DE is exact?

Well, we know that if My(x, y) and Nx(x, y) exist and are continuous and if

(⇤) M(x, y) dx + N(x, y) dy = 0

is exact with@F

@x(x, y) = M(x, y) and

@F

@y(x, y) = N(x, y),

then the mixed second-order partial derivatives Fxy and Fyx are continuous and

My = Fxy = Fyx = Nx.

ThenMy = Nx

is a necessary condition for (⇤) to be exact. As the following theorem states, itis also a su�cient condition.

Theorem (2 — Test for Exactness). Suppose that both M(x, y) andN(x, y) have continuous first-order partial derivatives in an open rectan-gular region R. Then

M(x, y) dx + N(x, y) dy = 0 is exact in R

if and only if

My = Nx holds at every point (x, y) 2 R.


Example.y3 dx + 3xy2 dy = 0

is exact sinceMy = 3y2 = Nx.

But if you divide thru by y2 to get

y dx + 3x dy = 0,

this is not exact sinceMy = 1 6= 3 = Nx.

Even though both DE’s have the same solution, xy3 = C or y = (Cx�1)1/3,one is exact and the other is not, so exactness is related to the precise form inwhich

M(x, y) dx + N(x, y) dy = 0

is written.


Method I for Solving Exact Equations

To solve an exact equation M(x, y) dx + N(x, y) dy = 0:

(1) Find a function F by integrating Fx = M with respect to x:

(⇤) F (x, y) =

ZM(x, y) dx + g(y).

(2) Find g(y) by taking the partial derivative of each side of (⇤) with respect toy and setting the result equal to N(x, y). Solve for g0(y). Then integrate g0(y)with respect to y to find g(y).

(3) Substitute g(y) into (⇤) to fully obtain F .

(4) Solutions of the DE are given implicitly by F (x, y) = C.

Example.(2xy � sec2 x)| {z }

M=?Fx

dx + (x2 + 2y)| {z }N=

?Fy

dy = 0.

My = 2x = Nx =) exact.

Thus M = Fx and N = Fy.

F (x, y) =

ZM dx + g(y) =

Z(2xy � sec2 x) dx + g(y) =)

F (x, y) = x2y � tan x + g(y).

Then Fy = x2 + g0(y) = x2 + 2y| {z }N

=)

g0(y) = 2y =) g(y) = y2.

ThusF (x, y) = x2y � tan x + y2 = C

is an implicit solution.


Method II for Solving Exact Equations

To solve an exact equation M(x, y) dx + N(x, y) dy = 0:

(1) Find a function F by integrating Fy = N with respect to y:

(⇤⇤) F (x, y) =

ZN(x, y) dy + h(x).

(2) Find h(x) by taking the partial derivative of each side of (⇤⇤) with respectto x and setting the result equal to M(x, y). Solve for h0(x). Then integrateh0(x) with respect to x to find h(x).

(3) Substitute h(x) into (⇤⇤) to fully obtain F .

(4) Solutions of the DE are given implicitly by F (x, y) = C.

Example.

(1 + exy + xexy)| {z }M=

?Fx

dx + (xex + 2)| {z }N=

?Fy

dy = 0.

My = ex + xex = Nx =) exact.

Thus M = Fx and N = Fy.

F (x, y) =

ZN dy + h(x) =

Z(xex + 2) dy + h(x) =)

F (x, y) = xexy + 2y + h(x).

Then Fx = exy + xexy + h0(x) = 1 + exy + xexy| {z }M

=)

h0(x) = 1 =) h(x) = x.

ThusF (x, y) = xexy + 2y + x = C



Example.⇣1

x+ 2y2x

⌘dx +

⇣2yx2 � cos y

⌘dy = 0, y(1) = ⇡.

My = 4xy = Nx =) exact.

F (x, y) =

Z ⇣2yx2 � cos y

⌘dy + h(x) =)

F (x, y) = x2y2 � sin y + h(x).

Fx = 2xy2 + h0(x) =1

x+ 2y2x =)

h0(x) =1

x=) h(x) = ln|x|.

ThusF (x, y) = x2y2 � sin y + ln|x| = C

is an implicit general solution. From y(1) = ⇡,

⇡2 � 0 + 0 = C =) C = ⇡2

ThenF (x, y) = x2y2 � sin y + ln|x| = ⇡2

is the particular solution.


Example.⇣y2 sin x

⌘dx +

⇣1

x� y

x

⌘dy = 0, y(⇡) = 1.

My = 2y sin x 6= � 1

x2+

y

x2= Nx =) not exact.

Try separable. x = 0 is not in domain of y, so multiply DE by x.

(y � 1) dy = xy2 sin x dx

Since y(⇡) = 1, y = 0 is not a constant solution.

Also, y = 1 is not a constant solution since it doesn’t satisfy the DE.

y � 1

y2dy = x sin x dx =)

Zy � 1

y2dy =

Zx sin x dx + K =)

x sin x+&

1 � cos x�&

0+ � sin x

Z ⇣1

y� 1

y2

⌘dy = �x cos x + sin x + K =)

ln|y| +1

y= �x cos x + sin x + K =)

From y(⇡) = 1,

0 + 1 = �⇡(�1) + 0 + K =) K = 1� ⇡.

Then

ln|y| +1

y= �x cos x + sin x + 1� ⇡


first-order di erential equationsfacstaff.cbu.edu/wschrein/media/m231 notes/m231c2.pdf18 2....

Documents