Chapter 2, Part 1

FIRST ORDER EQUATIONS

F (x, y, y′) = 0

Basic assumption: The equation

can be solved for y′; that is, the

equation can be written in the form

y′ = f(x, y) (1)

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2.1. First Order Linear Equa-

tions

Equation (1) is a linear equation if

f has the form

f(x, y) = P (x)y + q(x)

where P and q are continuous

functions on some interval I. Thus

y′ = P (x)y + q(x)

2

Standard form:

The standard form for a first order

linear equation is:

y′ + p(x)y = q(x)

where p and q are continuous

functions on the interval I

(Note: A differential equation which

is not linear is called nonlinear.)

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Examples:

1. Find the general solution:

y′ = 2y

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2. Find the general solution:

y′ + 2x y = 4x

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Solution Method:

Step 1. Establish that the equa-

tion is linear by writing it in standard

form

y′ + p(x)y = q(x).

Step 2. Multiply by e∫

p(x) dx :

e∫

p(x) dxy′+p(x)e∫

p(x) dxy = q(x)e∫

p(x) dx

[

e∫

p(x) dx y]′

= q(x)e∫

p(x) dx

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[

e∫

p(x) dx y]′

= q(x)e∫

p(x) dx

Step 3. Integrate:

e∫

p(x) dx y =∫

q(x)e∫

p(x) dx dx + C.

Step 4. Solve for y :

y = e−∫

p(x) dx∫

q(x)e∫

p(x) dx dx+Ce−∫

p(x) dx.

7

y = e−∫

p(x) dx∫

q(x)e∫

p(x) dx dx+Ce−∫

p(x) dx.

is the general solution of the equa-

tion.

Note: e∫

p(x) dx is called an inte-

grating factor

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2. Find the general solution:

xy′ + 3y =ln x

x2

y =ln x

x2− 1

x2+

C

x3

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3. Find the general solution:

x2y′ − 5xy = 3 − 6x3

y = 2x2 − 1

2x+ Cx5

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4. Find the general solution:

xy′ =2

√

x2 − 1− 2y + 2

y =2

√

x2 − 1

x2+ 1 +

C

x2

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5. Solve the initial-value problem:

y′+(cot x)y = 2cos x, y(π/2) = 3

y =5 − cos 2x

2 sin x12

6. Find the general solution:

y′ + 2x y = 2tan x

Answer:

y = e−x2∫

2ex2tan x dx + Ce−x2

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The term “linear:”

L[y] = y′ + p(x) y

is a linear operator:

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Set L[y] = y′ + p(x)y

L[y1 + y2] = (y1 + y2)′ + p(y1 + y2)

= y′1 + y′2 + py1 + py2

= y′1+py1+y′2+py2 = L[y1]+L[y2]

L[cy] = (cy)′ + p(cy) = cy′ + cpy

= c(y′ + py) = cL[y]

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Any ”operation” L that has the

two properties:

L[y1 + y2] = L[y1] + L[y2]

L[cy] = cL[y], c constant

is a linear operation.

1. Differentiation is a linear oper-

ation

2. Integration is a linear operation

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3. L[y] = y + p(x)y is a linear op-

eration.

2.2. Separable Equations

y′ = f(x, y)

is a separable equation if f has

the factored form

f(x, y) = p(x)h(y)

where p and h are continuous

functions. Thus

y′ = p(x)h(y)

is the ”standard form” of a separa-

ble equation.17

Example: Find the general solution

of:

y′ = xy2 + x

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Solution Method

Step 1. Establish that the equa-

tion is separable.

Step 2. Divide both sides by h(y)

to “separate” the variables.

1

h(y)y′ = p(x) or q(y)y′ = p(x)

which, in differential form, is:

q(y) dy = p(x) dx.

the variables are “separated.”

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Step 3. Integrate

∫

q(y) dy =∫

p(x) dx + C

Q(y) = P (x) + C

where Q′(y) = q(y), P ′(x) = p(x)

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Notes:

1. Q(y) = P (x)+C is the general

solution. Typically, this is an im-

plicit relation; you may or may not

be able to solve it for y.

2. h(y) = 0 is a source of singular

solutions: If k is a number such

that

h(k) = 0, then y = k

MIGHT be a singular solution.21

Examples:

1. Find the general solution:

y′ =2xy2 + 8x

4y

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2. Find the general solution:

dy

dx= 4x

√

y − 2 (Ch1, pg 31)

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3. Find the general solution:

dy

dx− xy2 = −x

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4. Find the general solution:

dy

dx=

ex−y

1 + ex

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5. Find the general solution and

any singular solutions:

(1 + x2 + y2 + x2y2)dy

dx= 2xy2

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2.3. Related Equations & Trans-

formations

A. Bernoulli equations

An equation of the form

y′ + p(x)y = q(x)yk, k 6= 0, 1

is called a Bernoulli equation.

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The change of variable

v = y1−k

transforms a Bernoulli equation into

v′ + (1 − k)p(x)v = (1 − k)q(x).

which has the form

v′ + P (x)v = Q(x),

a linear equation.

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Examples:

1. Find the general solution:

y′ − 4y = 2ex √y

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2. Find the general solution:

xy′ + y = 3x3 y2

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B. Homogeneous equations

y′ = f(x, y) (1)

is a homogeneous equation if

f(tx, ty) = f(x, y)

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If (1) is homogeneous, then the change

of dependent variable

y = vx, y′ = v + xv′

transforms (1) into a separable equa-

tion:

y′ = f(x, y) → v+xv′ = f(x, vx) = f(1, v)

which can be written

1

f(1, v) − vdv =

1

xdx;

the variables are separated.

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Examples:

1. Find the general solution:

y′ =x2 + y2

2xy

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2. Find the general solution:

dy

dx=

x2ey/x + y2

xy

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3. Find the general solution:

dy

dx=

y2

xy + y2

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