First-Order ODEs
Gyeongsang National University
Dept. of Information & Communication Engineering
Differential Equations?
The amount of salt after 𝑡 minutes? Inflow of salt water 10 l/min
100g of salt/1 Liter
10 l/min outflow Water 1000 l
𝑦(𝑡) : The amount of salt after 𝑡 minutes
𝑦′(𝑡) : Increase in salt at time 𝑡 (rate of change) 𝑦′ 𝑡 = 1000 − 0.01𝑦 𝑡
Basic Concept
• Differential equations
– An equation containing a function that differentiates an unknown function more than once
• Ordinary differential equations, ODE
– An equation consisting of functions that differentiate an unknown function 𝑦(𝑥) more than once
– Differential equations can include 𝑦(𝑥), a known function of 𝑥, and a constant
• 𝑛
𝑡ℎ-order ODE
– When the 𝑛𝑡ℎ-order derivative of 𝑦(𝑥) is the derivative of the highest order in the equation
– The order of the following differential equation?
• First-order ODE always includes 𝑦
′,
and can include
𝑦, 𝑥
• Implicit form of first-order ODE
• Explicit form of first-order ODE
• A funiction 𝑦 = ℎ 𝑥 is called of a given ODE:
– If ℎ 𝑥 is defined and differentiable on the open interval 𝑎 < 𝑥 < 𝑏 – If the equation becomes identity if 𝑦 and 𝑦′ are replaced with ℎ and ℎ′
• Example 1) Find the solution of 𝑦
′= cos 𝑥
Concept of Solution
• Example 2) Find the solution of 𝑦′ = 0.2𝑦
Concept of Solution
• A solution containing an arbitrary constant 𝑐 is called a general
solution of the ODE, e.g.,
𝑦 = 𝑐𝑒
0.2𝑡.
• If we choose a specific 𝑐 (e.g., 𝑐 = 1), the solution is called a
particular solution of the ODE.
– A particular solution does not contain any arbitrary constant.
• A particular solution is obtained from a general solution by an initial
condition
𝑦 𝑥
0= 𝑦
0with given value
𝑥
0and
𝑦
0.
– The solution curve should pass through the point (𝑥0, 𝑦0).
• An ODE, together with an initial condition, is called initial value
problem.
• Example 4)
• Step 1. The transition from the physical situation(physical system) to
is mathematical formulation(mathematical model)
• Step 2. The solution by a mathematical method
• Step 3. The physical interpretation of the result
Example 5) Given an amount of a radioactive substance
0.5𝑔, find the
amount present at time
𝑡?
Physical information) Experiments show that at each instant a
radioactive substance decomposes
– and is thus decaying in time –
proportional to the amount of substance present.
• Step 1. Setting up a mathematical model of the physical process
– 𝑦(𝑡) : the amount of substance present at any time 𝑡 – 𝑦′(𝑡) : the time rate of change of substance
– 𝑑𝑦
𝑑𝑡 = −𝑘𝑦 𝑘 > 0 , 𝑦 0 = 0.5
• Step 2. Mathematical solution
– 𝑦(𝑡) = 𝑐𝑒−𝑘𝑡 : general solution
– 𝑦(𝑡) = 0.5𝑒−𝑘𝑡 : particular solution
• Step 3. Interpretation of result
• Separable ODE: ODEs that can be solved by separating variables.
➢ First-order ODE
➢ Separable equation
➢ Separating variables
➢ Replace
1 𝐺(𝑦)to 𝑔(𝑦)
➢ Integrating both sides
• Example 1) Find a general solution of 𝑦
′= 1 + 𝑦
2Separable ODEs
Separating variables, Integrating, න 1 𝑎2 + 𝑦2 𝑑𝑦 → 1 𝑎tan −1 𝑥 𝑎 + 𝑐Separable ODEs
Separable ODEs
• Solving previous example by separating variables
– 𝑦′ = 0.2𝑦 – 𝑦′ = −0.2𝑦
• Example 4) Radiocarbon dating. A mummy was founded in ice. The ration of carbon 146
C
to 126C carbon 146C6
12C in this mummy is 52.5%. When did this
mummy approximately live and die?
– Physical information) In living organisms, the ratio of radioactive carbon 146C to
ordinary carbon 126C is constant. When an organism dies, the amount of 146C decreases by half every 5715 years. radioactive substance decomposes proportional to the amount of substance present.
Differential Equations?
The amount of salt after 𝑡 minutes? Inflow of salt water 10 l/min
100g of salt/1 Liter
10 l/min outflow Water 1000 l
• Example 6) Heating an office building. Suppose that in winter the daytime temperature in an office building, is maintained at 70°F. Heating is shup off at 10 PM and turned on again 6 AM. On a certain day the temperature inside the building at 2 AM was founded to be 65°F. The outside temperature was 45°F at 10 PM to 6 AM. What was the temperature inside the building when the heat was turned on at 6 AM?
– Physical information) Experiments show that the time rate of change of the temperature T of the object is proportional to the difference between T and the temperature of the surrounding medium (Newton’s law of cooling). Therefore, the rate of temperature change inside building is proportional to the temperature between inside and outside of building.
Extended Method:
Reduction to Separable Form
• Extended method: Certain non-separable ODEs can be separable
by transformations.
• Substitution to
𝑦Extended Method:
Reduction to Separable Form
Extended Method:
• Partial derivatives
From 𝜕𝑢 𝜕𝑥, 𝜕𝑢 𝜕𝑦, find 𝑢(𝑥, 𝑦)• From
𝜕𝑢 𝜕𝑥= 2𝑥𝑦
3+ 1,find
𝜕2𝑢 𝜕𝑥𝜕𝑦• From
𝜕𝑢 𝜕𝑦= 𝑥
23𝑦
2+ 1, find
𝜕2𝑢 𝜕𝑥𝜕𝑦Partial Derivatives
(By regarding y as a constant, differentiate u by x) (By regarding x as a constant, differentiate u by y)
• Total differential
Exact Differential Equation
Total differential of 𝑢 𝑥, 𝑦 = 𝑥 + 𝑥2𝑦3 = 𝑐:
General solution of the differential equation: 𝑢 𝑥, 𝑦 = 𝑥 + 𝑦2𝑦3 = 𝑐 𝑦′ = 𝑑𝑦
𝑑𝑥 = −
1 + 2𝑥𝑦3 3𝑥2𝑦2
• Definition of exact differential equation
– Transform first-order ODE 𝑀 𝑥, 𝑦 + 𝑁 𝑥, 𝑦 𝑦′ = 0 as following form:
– If 𝑀 𝑥, 𝑦 𝑑𝑥 + 𝑁 𝑥, 𝑦 𝑑𝑦 is in the form of 𝑑𝑢 = 𝜕𝑢
𝜕𝑥𝑑𝑥 + 𝜕𝑢
𝜕𝑦𝑑𝑦, we call this
equation as exact differential equation.
Exact Differential Equation
•
Determine whether it is an exact differential equation or not
If first-order ODE 𝑀 𝑥, 𝑦 𝑑𝑥 + 𝑁 𝑥, 𝑦 𝑑𝑦 = 0 become an exact DE, we have 𝜕𝑢 𝜕𝑥 = 𝑀, 𝜕𝑢 𝜕𝑦 = 𝑁, equivalently we have 𝜕2𝑢 𝜕𝑥𝜕𝑦 = 𝜕𝑀 𝜕𝑦, 𝜕2𝑢 𝜕𝑥𝜕𝑦 = 𝜕𝑁 𝜕𝑥. 𝜕𝑀 𝜕𝑁
• General solution of the exact DE
– From 𝜕𝑢
𝜕𝑥 = 𝑀, obtain 𝑢 = 𝑀𝑑𝑥 + 𝑘(𝑦) by integration
– From 𝑢 = 𝑀𝑑𝑥 + 𝑘(𝑦) , we obtain 𝜕𝑢𝜕𝑦 and then, obtain 𝑘(𝑦) by comparing with 𝑁(𝑥, 𝑦).
– From 𝑢 = 𝑁𝑑𝑥 + 𝑙(𝑥), we obtain 𝜕𝑢
𝜕𝑥 and then, obtain 𝑙(𝑥) by comparing
with 𝑀(𝑥, 𝑦).
➢ General solution has an implicit form as 𝑢 𝑥, 𝑦 = 𝑐.
• Example 1) find a solution of the following equation
cos 𝑥 + 𝑦 𝑑𝑥 + 3𝑦
2+ 2𝑦 + cos 𝑥 + 𝑦 𝑑𝑦 = 0
Exact Differential Equation
Step 1: determining whether the exact DE or not
• Example 2) Find the solution of 𝑥 − 𝑦 𝑑𝑥 − 𝑑𝑦 = 0.
• Example 3) Find the solution of −𝑦𝑑𝑥 + 𝑥𝑑𝑦 = 0
Exact Differential Equation
Step 1: determining whether the exact DE or not
This equation is not an exact DE but, if we multiply 1
𝑥2 to both sides,
it becomes the exact DE.
일반해: y
• Integrating factor
– For a non-exact DE:
– if a DE multiplied by function F(x, y) is an exact DE,
– We call F(x, y) as an integrating factor.
• Equation in example 3 has multiple integrating factors.
Integrating Factors
• Since the condition that 𝑀𝑑𝑥 + 𝑁𝑑𝑦 = 0 become an exact ODE is 𝜕𝑀
𝜕𝑦 = 𝜕𝑁
𝜕𝑥,
• the condition that 𝐹𝑃𝑑𝑥 + 𝐹𝑄𝑑𝑦 = 0 become an exact ODE is
• By the product rule, we have (𝐹𝑦 = 𝜕𝐹
𝜕𝑦, 𝐹𝑥 = 𝜕𝐹 𝜕𝑥)
• However, it is difficult to find the integrating factor to satisfy this condition.
Integrating Factors
• Finding the integrating factor that has a single variable (𝑥 or 𝑦). • Let 𝐹 be 𝐹 = 𝐹(𝑥), we have 𝐹𝑦 = 0, 𝐹𝑥 = 𝐹′ = 𝑑𝐹
𝑑𝑥, and then
– Dividing both sides by 𝐹𝑄,
– Theorem 1. if 𝑅 is a function of x only, the integrating factor is obtained by.
Integrating Factors
• Example 5) Solve the following differential equation:
• 단계 1) Determining whether the exact DE or not
• 단계 2) Finding the integrating factor and general solution
– R is not a function of x only, we cannot obtain integrating factor from Theorem 1.
– Thus, we obtain the integrating factor as 𝐹∗ 𝑦 = 𝑒−𝑦 and the equation
becomes an exact DE as
– The general solution is obtained as
– By the initial condition, the particular solution is obtained as
• Example 6) Find the general solution of 𝑥
2+ 𝑦
2𝑑𝑥 − 2𝑥𝑦𝑑𝑦 = 0
1.5 Linear ODEs. Bernoulii
• If the first-order ODE can be represented by following form, we call it
first-order linear ODE and the following form is called as standard
form
• For 𝑓 𝑥 𝑦
′+ 𝑝 𝑥 𝑦 = 𝑟(𝑥), it can be represented by the standard
form as
• Represent 𝑦
′cos 𝑥 + 𝑦 sin 𝑥 = 𝑥 as the standard form
• Determine whether the following DEs are linear or not.
• For the linear ODE with 𝑟 𝑥 = 0, we call it
homogeneous
linear
ODE.
• By separating variables,
• For the linear ODE with 𝑟(𝑥) ≠ 0, we call it
nonhomogeneous
linear
ODE.
• Transforming it to exact DE by multiplying the integrating factor
• Or we can obtaining the solution by following steps:
1. Multiplying an arbitrary 𝐹(𝑥) to the standard form equation:
2. Finding 𝐹(𝑥) that makes the left side becomes 𝐹𝑦 ′ = 𝐹′𝑦 + 𝐹𝑦′:
3. Substituting 𝐹 = 𝑒ℎ and ℎ′ = 𝑝 to the equation:
4. Integrating the equation:
5. Obtaining a general solution as
→ General solution of the equation:
Non-homogeneous Linear ODEs
• Example 1) Find a particular solution of 𝑦′ + 𝑦 tan 𝑥 = sin 2𝑥 , 𝑦 0 = 1
1. Transforming to the standard form
2. Finding ℎ
3. Finding 𝑒
ℎNon-homogeneous Linear ODEs
Non-homogeneous Linear ODEs
• Example 3) When initial current is 0 A, find 𝐼(𝑡) at time 𝑡.
Voltage drop: 𝑅𝐼(𝑡)
Bernoulli Equation
• Some of non-linear ODE can be transformed into linear ODE.
• Standard form of Bernoulli equation
• Example 4) Find a general solution of 𝑦
′= 𝐴𝑦 − 𝐵𝑦
2Bernoulli Equation
From 𝑢 𝑥 = 𝑦 𝑥 −1,