Engineering Mathematics
Chapter 2-1
Dr. Kwang-Bock You
Chapter 1: Introduction to Differential Equations
• A differential equation (DE) is an equation con
taining the derivatives of one or more depende
nt variables with respect to one or more indepe
ndent variables
• DEs can be classified by:
– Type
– Order
– Linearity
Chapter 1: Introduction to Differential Equations
Chapter 1: Introduction to Differential Equations
Chapter 1: Introduction to Differential Equations
■ General Solution and Particular Solution
1st-order ode:
y‘=cos x
General Solution:
y=sin x + c,
where c = (any) arbitrary constant
G( x, y, c1 , c2 ,K , cn ) 0 : (n-parameter) family of solution depending upon —
c“
A solution of a D.E. that is free of arbitrary parameter(s) is called a —
Particular Solution“,
such as y=sin x, y=sin x + 5, ….
■ Singular Solution => A solution that cannot be obtained by specializing any of the
parameters in the family of solutions. Such an extra solution is called a Singular Solution.
Ex.) y
xy1/ 2
y
( x 2 / 4 c) 2
y
x 4 /16
y
0
general Solution
Particular Solution, c
Singular Solution
0
§
§
§
§
§
§
Separable Equations
Linear Equations
Exact Equations
Solutions by Substitutions
Linear Models => CKT
Modeling /w Systems of 1st-Order Diff.
Equations => Networks
Chapter 2: First-Order Differential Equations
Separable Equations
Consider the first-order equation
𝒅𝒚
𝒅𝒙
𝒅𝒚
𝒅𝒙
Thus,
= 𝒇 𝒙, 𝒚
= 𝒈 𝒙 when 𝒇 𝒙, 𝒚 does not depend on
the variable 𝒚.
=> can be solved by integration
𝒚=𝑮 𝒙 +𝑪
where 𝑮 𝒙 is an antiderivative (indefinite integral)
of 𝒈 𝒙 .
Chapter 2: First-Order Differential Equations
Definition <Separable D.E.>
A differential equation is called separable if it
can be written
!
𝒅𝒚
𝒚 = 𝒅𝒙 = 𝒈 𝒙 𝒉(𝒚)
Chapter 2: First-Order Differential Equations
Example 1) Solving a Separable DE
Solve 𝟏 + 𝒙 𝒅𝒚 = 𝒚 𝒅𝒙
<= Separate variables
𝟏
𝟏
𝟏
𝟏
∴ 𝒚 𝒅𝒚 = 𝟏&𝒙 𝒅𝒙 → ∫ 𝒚 𝒅𝒚 = ∫ 𝟏&𝒙 𝒅𝒙
So, 𝒍𝒏 𝒚 = 𝒍𝒏 𝟏 + 𝒙 + 𝑪
∴𝒚=𝑪 𝟏+𝒙
Chapter 2: First-Order Differential Equations
Example 2) Solve
𝒅𝒚
𝒙
=
−
,
𝒚
𝟒
=
−𝟑
𝒅𝒙
𝒚
∴ 𝒚 𝒅𝒚 = −𝒙 𝒅𝒙 → H y 𝒅𝒚 = − H 𝒙 𝒅𝒙
So, 𝒙𝟐 + 𝒚𝟐 = 𝑪𝟐
Now, by using initial value, 𝒚 𝟒 = −𝟑,
Thus, the implicit solution; 𝒙𝟐 + 𝒚𝟐 = 𝟓𝟐
However, in the case of the explicit solution;
∴ 𝒚 = ± 𝟐𝟓 − 𝒙𝟐 → 𝒚 = − 𝟐𝟓 − 𝒙𝟐
Chapter 2: First-Order Differential Equations
■ Losing Solution
h( y )
y
0
dy
dx
y
r
r
g ( x ) h( y )
y
all 0
r (constant solution)
singular solution
Example 3 Losing a Solution
dy / dx
y2 4
Solution
dy
dx or
y2 4
y 2
ln
y 2
Note: y
1/4
y 2
4 x c2
1/ 4
dy
y 2
or
1
ln y 2
4
dx
y 2
y 2
e
4 x c2
.
y
1
y 2
4
1 ce4 x
2
1 ce4 x
x c1
(5)
(6)
2 (singular solution) /cannot be obtain from (6) for any choice of the parameter
Chapter 2: First-Order Differential Equations
Example 4 An Initial-Value Problem
cos x(e
dy
y)
dx
2y
e y sin 2 x, solve y (0)
Solution
e2 y
y
dy
y
e
(e y
e
y
ye y ) dy
ye
y
e
y(0) 0
e
y
ye
sin 2 x
dx.
cos x
y
e
y
y
2 sin x dx
2cos x c.
=> c
4
4 2cos x.
0
Chapter 2: First-Order Differential Equations
• Find a singular solution(s),
𝑑𝑦
= 𝑥 1 − 𝑦2
𝑑𝑥
(Sol.)
1
1 − 𝑦2
1 2
−1
∴ sin 𝑦 = 𝑥 + 𝑐
2
𝑑𝑦
𝑑𝑦 = 𝑥𝑑𝑥
𝑥2
→ 𝑦 = sin
+𝑐
2
• When 𝑦 = ±1 then 𝑑𝑥 = 0, that is, 𝑦 = ±1 are
singular solutions.
Chapter 2: First-Order Differential Equations
• Find solution(s),
𝑑𝑦
!
𝑒 𝑦
𝑑𝑥
= 𝑒 "# + 𝑒 "$!"#
Chapter 2: First-Order Differential Equations
• Solve
𝑑𝑦 𝑦 2 − 1
= 2
,
𝑑𝑥 𝑥 − 1
𝑦 2 =2
Chapter 2: First-Order Differential Equations
• Solve
𝑑𝑦
+ 2𝑦 = 1 ,
𝑑𝑡
5
𝑦 0 =
2