HANDOUT DIFFERENTIAL EQUATIONS For International Class
Nikenasih Binatari
NIP. 19841019 200812 2 005
Mathematics Educational Department Faculty of Mathematics and Natural Sciences State University of Yogyakarta 2010
CONTENTS
Cover Contents I. Introduction to Differential Equations - Some basic mathematical models………………………………… 1 - Definitions and terminology……………………………………… 3 - Classification of Differential Equation……………………………. 7 - Initial Value Problems…………………………………………….. 9 - Boundary Value Problems………………………………………... 10 - Autonomous Equation……………………………………………. 11 - Definition of Differential Equation Solutions……………………... 16
II. First Order Equations for which exact solutions are obtainable - Standart forms of first order Differential Equation………………. 19 - Exact Equation…………………………………………………….. 20 - Solution of exact differential equations…………………………... 22 - Method of grouping………………………………………………. 25 - Integrating factor………………………………………………….. 26 - Separable differential equations…………………………………... 28 - Homogeneous differential equations……………………………... 31 - Linear differential Equations……………………………………… 34 - Bernoulli differential equations…………………………………… 36 - Special integrating factor………………………………………….. 39 - Special transformations…………………………………………… 40
III. Application of first order equations - Orthogonal trajectories…………………………………………... 43 - Oblique trajectories………………………………………………. 46 - Application in mechanics problems and rate problems………….. 48
IV. Explicit methods of solving higher order linear differential equations - Basic theory of linear differential equations……………………… 51 - The homogeneous linear equation with constant coefficient……. 54 - The method of undetermined coefficients……………………….. 58 - Variation of parameters…………………………………………... 61
STATE UNIVERSITY OF-YOGYAKARTA FACULTY OF-MATHEMATICS AND NATURAL SCIENCE DEPARTMENT OF MATHEMATICS EDUCATION
First Meeting : I. Introduction to Differential Equations - Some Basic Mathematical Models
- Definitions and Terminology
The study of differential equations has attracted the attention of many of the world’s greatest mathematicians during the past three centuries. To give perspective to your study of differential equations, first, we use a problem to illustrate some of the basic ideas that we will return to and elaborate upon frequently throughout the remainder book.
I.1 Some Basic Mathematical Models
Consider the diagram below.
Express assumptions in Mathematical Assumptions terms of differential formulation equations If necessary, alter assumptions or increase
resolution of the model Solve the DEs
Check model Display model predictions, Obtain predictions with e.g., graphically solutions known facts
Example 1.1 The population of the city of Yogyakarta increases at a rate proportional to the number of its inhabitants present at any time t. If the population of Yogyakarta was 30,000 in 1970 and 35,000 in 1980, what will be the population of Yogyakarta in 1990?
We can use the theory of differential equation to solve this problem. Suppose 푦 푡 defines the number of population at time-t. Therefore, the first information can be transformed into mathematical model as 푑푦 = 퐾푦 푑푡 for some constant K, 푦 1970 = 30.000, and 푦 1980 = 35.000.
Just what is a differential equation and what does it signify? Where and how do differential equations originate and of what use are they? Confronted with a differential equation, what does one do with it, how does one do it, and what are the results of such activity? These questions indicate three major aspects of the subject: theory, method, and application. Therefore, things that will be discussed in this handout are how the forms of differential equations are, the method to find its solution for each form and the last one is its applications.
When we talk about something that is constantly changing, for example rate, then I can say, in truth, we talk about derivative. Many of the principles underlying the behavior of the natural world are statement or relations involving rates at which things happen. When expressed in mathematical terms the relations are equations and the rates are derivatives. Equations containing derivatives are differential equations.
I.2 Definitions and Terminology
Definition Differential Equation An equation containing the derivatives of one or more dependent variables, with respect to one or more independent variables, is said to be a differential equation (DE).
For examples
푑푦 1. = 2푥 푑푥 푑3푦 푑2푦 푑푦 2. − 2 − 4 + 8푦 = 0 푑푥3 푑푥2 푑푥
푑푦 2 3. − 2푦 = 0 푑푥 4. 푥2 + 푦2 푑푥 + 푦 − 푥 푑푦 = 0 푑푦 5. 푥 + 푥 + 1 푦 = 푥3 푑푥
Other examples of differential equation are 1. Radioactive Decay. The rate at which the nuclei of a substance decay is proportional to the amount (more precisely, the number of nuclei) 퐴(푡) of the substance remaining at time t. 푑퐴 = 푘퐴 푑푡 2. Newton’s Law of Cooling/warming. Suppose 푇(푡) represents the temperature of a body at time t, Tm the temperature of the surrounding medium. The rate at which the temperature of a body changes is proportional to the difference between the temperature of the body and the temperature of the surrounding medium. 푑푇 = 푘 푇 − 푇 푑푡 푚
More examples can be found at page 20 – 24, A first course in Differential Equation with modeling application, Dennis G. Zill.
Exercises. 1
Population Dynamics.
1. Assumed that the growth rate of a country at a certain time is proportional to the total population of the country at that time 푃(푡). Determine a differential equation for the population of a country when individuals are allowed to immigrate into the country at a constant rate 푟 > 0. How if individuals are allowed to emigrate from the country at a constant rate 푟 > 0. 2. In another model of a changing population of a community, it is assumed that the rate at which the population changes is a net rate – that is, the difference between the rate of births and the rate of deaths in the community. Determine a model for the population 푃(푡) if both the birth rate and the death rate are proportional to the population present at time t. 3. Using the concept of net rate introduced in Problem 2, determine a model for a population 푃(푡) if the birth rate is proportional to the population present at time t but the death rate is proportional to the square of the population present at time t. 4. Modify the model in problem 3 for the net rate at which the population 푃(푡) of a certain kind of fish changes by also assuming that the fish are harvested at a constant rate 푟 > 0.
Newton’s Law of Cooling/Warming
5. A cup of coffee cools according to Newtion’s Law of cooling. Use data from the graph of the temperature 푇 푡 in figure
beside to estimate the constants 푇푚 , 푇0 and k in a model of the form of a first-order initial-value problem 푑푇 = 푘 푇 − 푇 , 푇 0 = 푇 푑푡 푚 0
6. The ambient temperature 푇푚 could be a function of time t. Suppose that in an artificially controlled
environment, 푇푚 is periodic with a 24-hour period as illustrated in Figure beside. Devise a mathematical model for the temperature 푇(푡) of a body within this environment.
Spread of a Disease/Technology
7. Suppose a student carrying in a flu virus returns to an isolated college campus of 1000 students. Determine a differential equation for the number of people 푥(푡) who have contracted the flu if the rate at which the disease spreads is proportional to the number of interaction between the number of students who have the flu and the number of number of students who have the flu and the number of students who have not yet been exposed to it.
Mixtures
8. Supposes that a large mixing tank initially holds 300 gallons of water in which 50 pounds of salt have been dissolved. Pure water is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at the same rate. Determine the differential equation for amount of salt 퐴 푡 in the time at time t. What is 퐴 0 . 9. Suppose that a large mixing tank initially holds 300 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate 2 gal/min. If the concentration of the solution entering is 2 lb/gal, determine a differential equation for the amount of salt 퐴 푡 in the tank at time t.
Falling bodies and Air Resistance
10. For high-speed motion through the air – such as the skydiver shown in figure below, falling before the paracute is opened – air resistance is closer to a power of the instantaneous velocity 푣(푡). Determine a differential equation for the velocity 푣(푡) of a falling body of mass m if air resistance is proportional to the square of the instantaneous velocity.
STATE UNIVERSITY OF-YOGYAKARTA FACULTY OF-MATHEMATICS AND NATURAL SCIENCE DEPARTMENT OF MATHEMATICS EDUCATION
Second Meeting : I. Introduction to Differential Equations - Classification of Differential Equations
- Initial Value Problems - Boundary Value Problems
Differential equation is divided into two big classes; those are ordinary differential equation and partial differential equation. The different is located at the number of its independent variable.
Definition Ordinary Differential Equation A differential equation involving ordinary derivatives of one or more dependent variables with respect to a single independent variable is called an ordinary differential equation (ODE).
Definition Partial Differential Equation A differential equation involving partial derivatives of one or more dependent variables with respect to more than one independent variable is called a partial differential equation (PDE)
Definition Order The order of the highest ordered derivative involved in a differential equation is called the order of the differential equation
Definition Linear Ordinary Differential Equation A linear ordinary differential equation of order n, in the dependent variable y and the independent variable x, is an equation that is in, or can be expressed in, the form 푑푛 푦 푑푛−1푦 푑푦 푎 푥 + 푎 푥 + ⋯ + 푎 푥 + 푎 푥 푦 = 푏 푥 0 푑푥푛 1 푑푥푛−1 푛−1 푑푥 푛
Where 푎0 is not identically zero.
Definition Nonlinear Ordinary Differential Equation A nonlinear ordinary differential equation is an ordinary differential equation that is not linear.
As given from the previous materials, there are many terminologies for differential equations. Among those terminologies, differential equations are then classified by type, order and linearity.
Classification by Type Refers to the number of its independent variables, differential equation is divided into two types, ordinary differential equation and partial differential equation.
Classification by Order Consider to the highest derivative in the equation, differential equation can be classified by nth-order ordinary differential equations.
Classification by Linearity From its linearity, differential equation is classified into linear differential equation and nonlinear differential equation.
Classification of Differential Equation can be easily seen from the diagram below:
Ordinary DE Types Partial DE
Differential Order n-th DE Equation (DE)
Linear DE Linearity Nonlinear DE
For example
푑푦 1. = 2푦 푑푥 First order homogeneous linear ordinary differential equation
푑3푦 푑2푦 푑푦 2. − 2 − 4 + 8푦 = 0 푑푥3 푑푥2 푑푥 Third order homogeneous linear ordinary differential equation 3 푑2 푦 3. − 2푦 = 0 푑푥2 Second order homogeneous nonlinear ordinary differential equation 4. 푥2 + 푦2 푑푥 + 푦 − 푥 푑푦 = 0 first order
푑푦 5. 푥 + 푥 + 1 푦 = 푥3 푑푥 first order non homogeneous ordinary differential equation
Suppose that you are driving along a road and you want to predict where you will be in the future. You need to know two things-- How fast you are driving. Where you start. Now suppose that you are the mayor of a city and want to predict how many students will be enrolled the schools in the future. Once again you need to know two things-- How fast the school population is changing. The current school population. The two parts of each of the problems above together make up an initial value problem or IVP. An IVP has two parts-- A differential equation describing how things change. One or more initial conditions describing how things start.
Definition of Initial Value Problems
On some interval I containing x0, the problem Solve :
푑푛 푦 = 푓 푥, 푦, 푦′ , … , 푦 푛−1 푑푥푛
Subject to : 푦 푥0 = 푦0, ′ 푛−1 푦 푥0 = 푦1, … , 푦 푥0 = 푦푛−1
Where 푦0, 푦1, 푦2, … 푦푛−1 are arbitrarily specified real constants, is called an initial value problem (IVP). The given values of the unknown function ′ 푛−1 and its first derivatives at a single point x0, 푦 푥0 = 푦0, 푦 푥0 = 푦1, … , 푦 푥0 = 푦푛−1 are called initial conditions.
Another type of problem consists of solving a linear differential equation in which the dependent variable 푦 or its derivatives are specified at different points. This type of differential equation is called boundary value problem (BVP)
STATE UNIVERSITY OF-YOGYAKARTA FACULTY OF-MATHEMATICS AND NATURAL SCIENCE DEPARTMENT OF MATHEMATICS EDUCATION
Third Meeting : I. Introduction to Differential Equations - Autonomous Equation
An important class of first order equations is those in which the independent variable does not appear explicitly. Such equations are called autonomous and have the form 푑푦 = 푓 푦 푑푡 For example
푑푦 = 푎푦 + 푏, 푑푡 where a and b are constants. Our goal now is to obtain important qualitative information directly from the differential equation, without solving the equation.
풅풚 Steps to draw the solution curves of = 풇 풚 . 풅풕
1. Find the equilibrium solutions, those are constant solutions or solutions which
푑푦 satisfying = 0. 푑푡 2. Determine which area such that the curve will increase or decrease.
푑푦 The curve will increase for area where > 0 and decrease for area where 푑푡 푑푦 < 0 푑푡 3. To investigate one step further, we can determine the concavity of the
푑2푦 solution curve by using its second derivative, 푑푡2 푑2푦 푑푦 = 푓′ 푦 = 푓′ 푦 푓 푦 푑푡2 푑푡 푑2푦 The graph of y versus t is concave up when > 0 while concave down 푑푡2 푑2푦 when < 0. 푑푡2
Look at the following figures.
Example 1.3
The population of the city of Yogyakarta increases at a rate proportional to the number of its inhabitants present at any time t. The population of Yogyakarta was 30,000 in 1970 and 35,000 in 1980. Determine the behavior of the number of its population.
Answer : As we modeled this problem before, suppose 푦 푡 defines the number of population at time- t. Therefore, the first information can be transformed into mathematical model as 푑푦 = 퐾푦 푑푡 for some constant K > 0, 푦 1970 = 30.000, and 푦 1980 = 35.000.
From the concept of derivative, we know that first derivative of a function is the rate. As we
푑푦 see that k > 0 and y > 0 which are implied that > 0. Therefore we can simply conclude 푑푡 that the number of population will increase all over the time.
Figure : y versus t for dy/dt = Ky.
Example I.4
If the population rate can be modeled into
푑푦 = 3 − 푦 푦 푑푡
Answer :
1. The equilibrium solutions are 푦 such that 3 − 푦 푦 ↔ 푦 = 3 or 푦 = 0 2. Here, we can analyze that for 0 < 푦 < 3 the derivative is positive valued while for 푦 < 0 or 푦 > 3 the derivative is negative valued. Therefore, the population will increase for 0 < 푦 < 3, while for 푦 < 0 or 푦 > 3 the population will decrease. 3. Determine the concavity. 푑2푦 푑푦 = 푓′ 푦 = 푓′ 푦 푓 푦 푑푡2 푑푡 푑2푦 Suppose 푓 푦 = 3 − 푦 푦, then 푓′ 푦 = −2푦 + 3. It’s implied that = 푑푡2 3 − 2푦 3 − 푦 푦.
푑푦 Figure : 푦 versus 푡 for = 3 − 푦 푦 푑푡
Definition
Consider the value for the solution 푦 that starts near the equilibrium solution 푦0 as 푡 → ∞. If
lim 푦 푡 = 푦0 푡→∞ then 푦0 is called an asymptotically stable solution. Otherwise, it’s called an unstable equilibrium solution.
Example I.4 above, 푦 = 3 is an asymptotically stable solution whereas 푦 = 0 is an unstable equilrium.
Exercises 3.
In each problem below, sketch the graph of y versus t, determine the critical (equilibrium) points, and classify each one as asymptotically stable or unstable.
푑푦 1. = 2푦 + 3푦2, 푦 0 = 3 푑푡 푑푦 2. = 푦 푦 − 1 푦 − 2 , 푦 0 = 4 푑푡 푑푦 3. = 푦 푦 − 1 푦 − 2 , 푦 0 = 1,2 푑푡 푑푦 4. = 푦 푦 − 1 푦 − 2 , 푦 0 = −1 푑푡 푑푦 5. = 푒푦 − 1, 푦 0 = 1 푑푡
STATE UNIVERSITY OF-YOGYAKARTA FACULTY OF-MATHEMATICS AND NATURAL SCIENCE DEPARTMENT OF MATHEMATICS EDUCATION
Fourth Meeting : I. Introduction to Differential Equations - Differential Equations Solution
As mentioned before, one of our goals in this course is to solve or to find the solutions of differential equations.
Definition Solution of a Differential Equation
Any function 푓 defined on some interval I, which when substituted into a differential equation reduces equation to an identity, is said to be a solution of the equation on the interval. In other words, 푓 is said to be a solution if it possess at least n derivatives and satisfies the differential equation, 푦 = 푓 푥 .
Consider the following differential equation
푑푦 푥 = − 푑푥 푦
Explicit solution
A solution in which the dependent variable is expressed in terms of the independent variable and constant is said to be an explicit solution. Specifically, an explicit solution of a differential equation that is identically zero on an interval I is said to be a trivial solution.
Example : 푦 = ± 4 − 푥2 is an explicit solution of differential equation above.
Implicit solution
A relation 퐺 푥, 푦 = 0 is said to be an implicit solution of an ordinary differential equation on an interval I provided there exists at least one function ∅ that satisfies the relation as well as the differential equation on I. in other words, 퐺 푥, 푦 = 0 defines the function ∅ implicitly.
Example : 푥2 + 푦2 − 4 = 0 is an implicit solution of differential equation above.
General solution
A solution containing single or multiple arbitrary constant is called general solution.
Example : 푥2 + 푦2 − 푐 = 0 for every constan 푐 ≥ 0 is general solution of differential equation above.
Particular solution
A solution of a differential equation that is free of arbitrary parameters is called a particular solution.
Example : suppose 푦 0 = 4 then 푐 = 16, therefore 푥2 + 푦2 − 16 = 0 is particular solution of differential equation where 푦 0 = 4.
Exercises 4.
1. Show that 푦 = 4푒2푥 + 2푒−3푥 is a solution of differential equation as follows 푑2푦 푑푦 + − 6푦 = 0 푑푥2 푑푥 푑푦 2. Show that 푦 = 푥2 + 푐 푒−푥 is the solution of + 푦 = 2푥푒−푥 . Given that 푦 0 = 2, 푑푥 find the particular solution. 푑2푦 푑푦 3. Show that 푦 = 푐 푒4푥 + 푐 푒−3푥 is the solution of − − 12푦 = 0 and then find the 1 2 푑푥2 푑푥 particular solution if it’s known that 푦 0 = 5 and 푦′ 0 = 6.
4. Show that for some choice of the arbitrary constants 푐1 and 푐2, the form 푦 = 푐1 sin 푥 +
푑2푦 푑푦 푐 cos 푥 is the solution of + = 0 but the boundary problems 푦 0 = 0 and 2 푑푥2 푑푥 푦 휋 = 1 doesn’t possess the solution. 2 3 5. Find the value of 푐1, 푐2 and 푐3 such that 푦 = 푐1푥 + 푐2푥 + 푐3푥 is the solution of 푑3푦 푑2푦 푑푦 푥3 − 3푥2 + 6푥 − 6푦 = 0 푑푥3 푑푥2 푑푥 That satisfying three conditions 푦 2 = 0, 푦′ 2 = 2, 푦"(2) = 6
STATE UNIVERSITY OF-YOGYAKARTA FACULTY OF-MATHEMATICS AND NATURAL SCIENCE DEPARTMENT OF MATHEMATICS EDUCATION
Fifth Meeting : II. First Order Differential Equations for Which Exact Solutions are obtainable - Standard forms of First Order Differential Equations - Exact Equation
In this chapter, we will discuss the solution of first order differential equation. The main idea is to check whether it is an exact or nonexact. For Nonexact first order differential equation, we should use integrating factor to transform nonexact differential equation into the new differential equation which is exact.
II.A Standard forms of First Order Differential Equations
First order differential equation can be expressed as derivative form 푑푦 = 푓 푥, 푦 … … (2.1) 푑푥 or as differential form
푀 푥, 푦 푑푥 + 푁 푥, 푦 푑푦 = 0 … … (2.2)
Definition 2.1 Let F be a function of two real variables such that F has continuous first partial derivatives in a domain D. The total differential dF of the function F is defined by the formula 휕퐹(푥, 푦) 휕퐹(푥, 푦) 푑퐹 푥, 푦 = 푑푥 + 푑푦 휕푥 휕푦 For all 푥, 푦 ∈ 퐷.
Definition 2.2 In a domain D, if there exists a function F of two real variables such that expression 푀 푥, 푦 푑푥 + 푁 푥, 푦 푑푦 equals to the total differential 푑퐹(푥, 푦) for all 푥, 푦 ∈ 퐷 then this expression is called an exact differential.
In other words, the expression above is an exact differential in D if there exists a function F such that
휕퐹(푥,푦) 휕퐹(푥,푦) = 푀(푥, 푦) and = 푁(푥, 푦) 휕푥 휕푦 for all 푥, 푦 ∈ 퐷.
EXACT EQUATION If 푀 푥, 푦 푑푥 + 푁 푥, 푦 푑푦 is an exact differential, then the differential equation 푀 푥, 푦 푑푥 + 푁 푥, 푦 푑푦 = 0 is called an exact differential equation.
The following theorem is used to identify whether a differential equation is an exact or not.
Theorem
Consider the differential equation 푀 푥, 푦 푑푥 + 푁 푥, 푦 푑푦 = 0 Where M and N have continuous first partial derivatives at all points (푥, 푦) in a rectangular domain D.
1. If the differential equation is exact in D, then
휕푀 푥, 푦 휕푁(푥, 푦) = 휕푦 휕푥 For all 푥, 푦 ∈ 퐷 2. Conversely, if 휕푀 푥, 푦 휕푁(푥, 푦) = 휕푦 휕푥 For all 푥, 푦 ∈ 퐷, then the differential equation is exact in D.
Example 2.1
Determine the following differential equation is exact or nonexact.
푑푦 2푥푦 − 9푥2 + 2푦 + 푥2 + 1 = 0 푑푥 Answer :
The differential equation is expressed in derivative form. Transformed into differential form, we get
2푥푦 − 9푥2 푑푥 2푦 + 푥2 + 1 푑푦 = 0
Suppose 푀 푥, 푦 = 2푥푦 − 9푥2 and 푁 푥, 푦 = 2푦 + 푥2 + 1, then the differential equation above is exact if and only if
휕푀 푥, 푦 휕푁 푥, 푦 = . 휕푦 휕푥
휕푀 푥, 푦 휕 2푥푦 − 9푥2 = = 2푥 휕푦 휕푦
While
휕푁 푥, 푦 휕푥 2푦 + 푥2 + 1 = = 2푥 휕푥 휕푥
휕푀 푥,푦 휕푁 푥,푦 Because = , then the differential equation above is an exact. 휕푦 휕푥
Exercises 5.
Determine whether it is an exact differential equation or not. Explained! 푥 1. 1 + ln 푥푦 푑푥 + 푑푦 = 0 푦 2. 푥2푦 푑푥 − 푥푦2 + 푦3 푑푦 = 0 3. 푦 + 3푥2 푑푥 + 푥 푑푦 = 0 4. cos 푥푦 − 푥푦 sin 푥푦 푑푥 − 푥2 sin 푥푦 푑푦 = 0 5. 푦푒푥푦 푑푥 + 2푦 − 푥푒푥푦 푑푦 = 0 More exercises can be found in the references.
STATE UNIVERSITY OF-YOGYAKARTA FACULTY OF-MATHEMATICS AND NATURAL SCIENCE DEPARTMENT OF MATHEMATICS EDUCATION
Sixth Meeting : II. First Order Differential Equations for Which Exact Solutions are obtainable - Solution of Exact Differential Equations
By using this theorem, the solution for exact differential equation above is a function F such that satisfies
휕퐹(푥,푦) 휕퐹(푥,푦) = 푀(푥, 푦) and = 푁(푥, 푦) 휕푥 휕푦 for all 푥, 푦 ∈ 퐷. Therefore,
푀 푥, 푦 푑푥 + 푁 푥, 푦 푑푦 = 0 휕퐹(푥, 푦) 휕퐹(푥, 푦) 푑푥 + 푑푦 휕푥 휕푦
푑퐹 푥, 푦 = 0
If we manipulate it for several operation, we obtain the solution can be expressed into
휕푀 푥, 푦 퐹 푥, 푦 = 푀 푥, 푦 휕푥 + 푁 푥, 푦 − 휕푥 푑푦 휕푦
Example 2.2
Solve the differential equation below
3푥 + 2푦 푑푥 + 2푥 + 푦 푑푦 = 0
Answer :
First we have to figure it out whether it’s an exact differential equation or not.
Suppose 푀 푥, 푦 = 3푥 + 2푦 and 푁 푥, 푦 = 2푥 + 푦
Therefore, we get 휕푀(푥, 푦) 휕푁(푥, 푦) = 2 푎푛푑 = 2 휕푦 휕푥
휕푀(푥,푦) 휕푁(푥,푦) Because = , then the differential equation above is an exact differential 휕푦 휕푥 equation.
So, there exist a function F such that satisfies
휕퐹(푥,푦) 휕퐹(푥,푦) = 푀(푥, 푦) and = 푁(푥, 푦) 휕푥 휕푦 We have
휕퐹(푥, 푦) = 푀 푥, 푦 = 3푥 + 2푦 휕푥 Then
3 퐹 푥, 푦 = 3푥 + 2푦 휕푥 + 휃 푦 = 푥2 + 2푦 + 휃 푦 2
Derived it for y,
휕퐹(푥, 푦) = 2 + 휃′ 푦 휕푦
휕퐹(푥,푦) Because = 푁(푥, 푦), then 휕푦
2 + 휃′ 푦 = 2 → 휃′ 푦 = 0
Here we get,
휃 푦 = 휃′ 푦 푑푦 = 0 푑푦 = 푐
From the operations above, we obtain the solution for the differential equation is
3 퐹 푥, 푦 = 푥2 + 2푦 + 푐 2
Exercise 6.
1. Solve the exact differential equations in Exersice 5. 2. Solve the following differential equations. a. 2푥푦 푑푥 + 푥2 + 1 푑푦 = 0 b. 푦2 + cos 푥 푑푥 + 2푥푦 + sin 푦 푑푦 = 0 c. 4푒2푥 + 2푥푦 − 푦2 푑푥 + 푥 − 푦 2 푑푦 = 0
STATE UNIVERSITY OF-YOGYAKARTA FACULTY OF-MATHEMATICS AND NATURAL SCIENCE DEPARTMENT OF MATHEMATICS EDUCATION
Seventh Meeting : II. First Order Differential Equations for Which Exact Solutions are obtainable - Method of Grouping - Integrating Factor Another way to find the solution of differential equation is by method of grouping that is grouping the terms of differential equation in such a way that its left member appears as the sum of certain exact differentials.
Example
2푥 cos 푦 + 3푥2푦 푑푥 + 푥3 − 푥2 sin 푦 − 푦 푑푦 = 0
Answer :
We can rewrite the differential equation above into the following term
2푥 cos 푦 푑푥 − 푥2 sin 푦 푑푦 + 3푥2푦 푑푥 + 푥3푑푦 − 푦푑푦 = 0
By using total difference for each term under the blankets, here we get,
1 푑 푥2 cos 푦 + 푑 푥3푦 − 푑 푦2 = 0 2
Integrating both sides, we have
1 푥2 cos 푦 + 푥3푦 − 푦2 = 0 2
1 So, the solution of the differential above is 푥2 cos 푦 + 푥3푦 − 푦2 = 0. 2
Exercise 7.1
Solve the following equations by method of grouping.
1. 3푥 + 2푦 푑푥 + 2푥 + 푦 푑푦 = 0 2. 푦2 + 3 푑푥 + 2푥푦 − 4 푑푦 = 0 3. 6푥푦 + 2푦2 − 5 푑푥 − 푥3 + 푦 푑푦 = 0 4. 푎2 + 1 cos 푟 푑푟 + 2푎 sin 푟 푑푎 = 0 5. 푦 sec2 푥 + sec 푥 tan 푥 푑푥 + tan 푥 + 2푦 푑푦 = 0
From the previous meeting, it has been discussed differential equation which is exact. How if the equation isn’t exact? Here, we introduced integrating factor that is a factor when multiplied into the equation, it will transform the equation into exact.
Definition If the differential equation 푀(푥, 푦) 푑푥 + 푁(푥, 푦) 푑푦 = 0 is not exact in a domain D but the differential equation 휇 푥, 푦 푀 푥, 푦 푑푥 + 휇 푥, 푦 푁 푥, 푦 푑푦 = 0 is exact in D, then 휇(푥, 푦) is called an integrating factor of the differential equation.
Note : Solution of the first equation is equals with the second equation.
Exercises 7.2
1. Consider the differential equation 4푥 + 3푦2 푑푥 + 2푥푦 푑푦 = 0 a. Show that this equation is not exact b. Find the integratiiing factor of the form 푥푛 , where n is a positive integer. c. Multiply the given equation through by the integrating factor found in (b) and solve the resulting exact equation. 2. Consider the differential equation 푦2 + 2푥푦 푑푥 − 푥62 푑푦 = 0 a. Show that this equation is not exact. b. Multiply the given equatttion through by 푦푛, where n is an integer and then determine n so that 푦푛 is an integrating factor of the given equation. c. Multiply the given equation through by the integrating factor found in (b) and solve the resulting exact equation. d. Show that y = 0 is a solution of the original nonexact equation but is not a solution of the essentially equivalent exact equation found in step (c). e. Graph several integral curves of the original equation, including all those whose equations are (or can be writen) in some special form.
STATE UNIVERSITY OF-YOGYAKARTA FACULTY OF-MATHEMATICS AND NATURAL SCIENCE DEPARTMENT OF MATHEMATICS EDUCATION
Eighth Meeting : II. First Order Differential Equations for Which Exact Solutions are obtainable - Separable Differential Equations
Next, we will discuss about some kind of nonexact differential equation and how it is solved.
A. SEPARABLE EQUATIONS Consider to the general form of differential equation in differential form. The differential equation is called separable differential equations if 푓 or M and N can be expressed into multiplication of function of each variable. In other words, M and N can be expressed into formula in the definition as follows :
Definition
An equation of the form
퐹 푥 퐺 푦 푑푥 + 푓 푥 푔 푦 푑푦 = 0
or
푑푦 = 푔 푥 푕 푦 푑푥 is called an equation with variables separable or simply a separable equation.
As we can see that separable equation which is expressed in differential form, in general, is 1 not exact, but it possesses an obvious integrating factor, namely where 푓 푥 ≠ 0 푓 푥 퐺 푦 1 and 퐺 푦1 ≠ 0 for some 푥1 and 푦1. Therefore, if we multiply this equation by this expression, we separate the variables, reducing it to the essentially equivalent equation 퐹 푥 푔 푦 푑푥 + 푑푦 = 0 푓 푥 퐺 푦
This last equation is exact since
휕 퐹 푥 휕 푔 푦 = 0 = 휕푦 푓 푥 휕푥 퐺 푦
The solution of the separable equation then simply can be resulted by integrating each variable
퐹 푥 푔 푦 푑푥 = 푑푦 푓 푥 퐺 푦 If there exists 푥1 and or 푦1 such that 푓 푥1 = 퐺 푦1 = 0, then we must determine whether any of these solutions of the original equation which were lost in the formal separation process.
Example.
Solve
푥푦4 푑푥 + 푦2 + 2 푒−3푥 푑푦 = 0
Answer :
This equation is an exact differential equation. We define the integrating factor of its equation is
1 , 푦4 ≠ 0. 푦4푒−3푥
Here, we obtain
푥 푦2 + 2 푑푥 + 푑푦 = 0 푒−3푥 푦4
푥 푦2 + 2 푑푥 = − 푑푦 푒−3푥 푦4
Using integration by parts
푥 푦2 + 2 푑푥 = − 푑푦 푒−3푥 푦4 yields
1 2 1 1 − + + − = 푐 푥 푥2 푦 푦3
Then, we have to check whether 푦 = 0 is a solution from its origin equation which is loss in separation process or not.
푑푦 푥푦4 푥푦4 푑푥 + 푦2 + 2 푒−3푥 푑푦 = 0 → = − 푑푥 푦2 + 2 푒−3푥
Here, we observe that 푦 = 0 is also a solution.
Exercise 8. Solve the differential equations below 1. 4푦 + 푦푥62 푑푦 − 2푥 + 푥푦2 푑푥 = 0 2. 1 + 푥62 + 푦2 + 푥2푦2 푑푦 = 푦2푑푥 3. 2푦 푥 + 1 푑푦 = 푥푑푥 4. 푥2푦2푑푦 = 푦 + 1 푑푥 5. 푒푦 sin 2푥 푑푥 + cos 푥 푒2푦 − 푦 푑푦 = 0
STATE UNIVERSITY OF-YOGYAKARTA FACULTY OF-MATHEMATICS AND NATURAL SCIENCE DEPARTMENT OF MATHEMATICS EDUCATION
Ninth Meeting : II. First Order Differential Equations for Which Exact Solutions are obtainable - Homogeneous Differential Equations
B. HOMOGENEOUS EQUATIONS Definition
A function 푓 of two variables x and y is said homogeneous if it satisfies
푓 푡푥, 푡푦 = 푓 푥, 푦
for any number t.
Example
푥2 + 푦2 푓 푥, 푦 = 푥푦 is a homogeneous function because for every number t,
푡푥 2 + (푡푦)2 푡2(푥2 + 푦2) 푥2 + 푦2 푓 푡푥, 푡푦 = = = = 푓 푥, 푦 푡푥 (푡푦) 푡2푥푦 푥푦
Definition
A function M of two variables x and y is said to be homogenous of degree-n if it satisfies
푀 푡푥, 푡푦 = 푡푛 푀 푥, 푦
Example
푀 푥, 푦 = 푥3 + 푥2푦 is a homogenous of degree-3 function because for every number t,
푀 푡푥, 푡푦 = 푡푥 3 + 푡푥 2 푡푦 = 푡3푥3 + 푡3푥2푦 = 푡3 푥3 + 푥2푦 = 푡3푀(푥, 푦) By here, we can define homogeneous differential equation as follows
Definition
A first differential equation
푑푦 푀 푥, 푦 푑푥 + 푁 푥, 푦 푑푦 = 0 or = 푓(푥, 푦) 푑푥 is called homogeneous differential equation if M and N are homogeneous functions of the same degree n or f is a homogenous function.
To solve homogeneous differential equation, we can use the following theorem:
Theorem
If
푀 푥, 푦 푑푥 + 푁 푥, 푦 푑푦 = 0
is a homogeneous differential equation, then the change of variables 푦 = 푣푥 transforms it into a separable differential equation in the variables v and x.
Here, we can summarize the steps of solving homogeneous differential equation as follow: 1. Recognize that the equation is a homogeneous differential equation 2. Transforms it into a separable differential equation by 푦 = 푣푥 3. Solve the separable differential equation in the variable v and x by integrating by parts. 푦 4. Inverse it into the origin differential equation 푣 = . 푥
Example.
Solve this differential equation
2푥푦 푑푥 + 푥2 + 1 푑푦 = 0
Answer:
As we can see that M and N are homogeneous functions with the same degree 2, therefore the we should transform it into separable equation by substituting 푦 = 푣푥. Here we get, 푑푦 = 푣푑푥 + 푥푑푣. Substitute into the equation, we get
2푥(푣푥) 푑푥 + 푥2 + 1 푣 푑푥 + 푥 푑푣 = 0
2푥2푣 푑푥 + 푥2 + 1 푣 푑푥 + 푥3 + 푥 푑푣 = 0
3푥2 + 1 푣 푑푥 + 푥3 + 푥 푑푣 = 0
The last equation is separable equation. To solve this equation, we multiply it with integrating 1 factor . We get, 푣 푥3+푥
3푥2 + 1 1 푑푥 + 푑푣 = 0 푥3 + 푥 푣 By integrating it in both side, we get
3푥2 + 1 1 푑푥 + 푑푣 = ln 푐 푥3 + 푥 푣
ln 푥3 + 푥 + ln 푣 = ln 푐
ln 푥3 + 푥 푣 = ln 푐
푥3 + 푥 푦 푥3 + 푥 푣 = 푐 → = 푐 → 푥2푦 + 푦 = 푐 푥 So, the solution of the differential equation above is 푥2푦 + 푦 = 푐.
Exercise 9.
Solve the following differential equations. 1. 푥 + 2푦 푑푥 + 2푥 − 푦 푑푦 = 0 2. 3푥 − 푦 푑푥 − 푥 + 푦 푑푦 = 0 3. 2푥푦 + 3푦2 푑푥 − 2푥푦 + 푥2 푑푦 = 0 4. 푦3푑푥 + 푥3 − 푥푦2 푑푦 = 0 5. 2푠2 + 2푠푡 + 푡2 푑푠 + 푠2 + 2푠푡 − 푡2 푑푡 = 0
STATE UNIVERSITY OF-YOGYAKARTA FACULTY OF-MATHEMATICS AND NATURAL SCIENCE DEPARTMENT OF MATHEMATICS EDUCATION
Tenth Meeting : II. First Order Differential Equations for Which Exact Solutions are obtainable - Linear Differential Equations
C. LINEAR EQUATIONS
Definition
A first-order ordinary differential equation is linear in the dependent variable y and the independent variable x if it is, or can be, written in the form
푑푦 + 푃 푥 푦 = 푄 푥 푑푥 By several multiplications, equation above can also be expressed as
푃 푥 푦 − 푄 푥 푑푥 + 푑푦 = 0
Suppose the integrating factor
휇 푥 = 푒 푃 푥 푑푥
The general solution of linear differential equation as follows
푦 = 휇−1 푥 휇 푥 푄 푥 푑푥 + 푐
Steps to solve linear differential equation are
1. If the differential equation is given as 푑푦 푎 푥 + 푏 푥 푦 = 푐 푥 푑푥 Rewrite it in the form 푑푦 + 푝 푥 푦 = 푞 푥 푑푥
2. Find the integrating factor 휇 푥 = 푒 푃 푥 푑푥
3. Evaluate the integral 휇 푥 푄 푥 푑푥 4. Write down the general solution 푦 = 휇−1 푥 휇 푥 푄 푥 푑푥 + 푐
5. If it is given an initial value, use it to find the constant C.
Example. Solve
푑푦 − 3푦 = 0 푑푥 Answer : 1. As we can see from the formula above that 푃 푥 = −3 and 푄 푥 = 0. Therefore, the integrating factor of the equation is
휇 푥 = 푒 −3 푑푥 = 푒−3푥 2. Multiply both sides with the integrating factor, yields 푑푦 푒−3푥 − 3푒−3푥 푦 = 0 푑푥 3. Evaluate
−3푥 휇 푥 푄 푥 푑푥 = 푒 . 0 푑푥 = 푐1
4. The general solution is
푦 = 휇−1 푥 휇 푥 푄 푥 푑푥 + 푐
3푥 = 푒 푐1 + 푐 3푥 푦 = 푐0푒
Exercise 10.
푑푦 3푦 1. + = 6푥2 푑푥 푥 푑푦 2. + 3푦 = 3푥2푒−푥 푑푥 푑푦 3. + 4푥푦 = 8푥 푑푥 푑푦 4. 푥 + 3푥 + 1 푦 = 푒−3푥 푑푥 푑푦 5. 푥 + 1 + 푥 + 2 푦 = 2푥푒−푥 푑푥
STATE UNIVERSITY OF-YOGYAKARTA FACULTY OF-MATHEMATICS AND NATURAL SCIENCE DEPARTMENT OF MATHEMATICS EDUCATION
Eleventh and II. First Order Differential Equations for Which Exact Solutions are obtainable twelfth Meetings - Bernoulli Differential Equations
D. BERNOULLI EQUATIONS
Definition
An equation of the form
푑푦 + 푃 푥 푦 = 푄 푥 푦푛 푑푥 is called a Bernoulli differential equation.
To solve Bernoulli differential equation, where n ≠ 0 or n ≠ 1, we can use the following theorem.
Theorem
Suppose n ≠ 0 or n ≠ 1, then the transformation 푣 = 푦1−푛 reduces the Bernoulli equation 푑푦 + 푃 푥 푦 = 푄 푥 푦푛 푑푥
into a linear equation in v.
Therefore, the steps to solve Bernoulli Differential Equation are : 1. Recognize that the differential equation is a Bernoulli Differential Equation. Then find the parameter n from the equation. 2. The main idea is to transform the Bernoulli Differential Equation into Linear Differential Equation, therefore the things that we should do are a. Divided both side with 푦푛.
푑푣 b. Transform the equation by 푣 = 푦1−푛 . Here we get, = 1 − 푛 푦−푛. 푑푦 c. By using manipulations, we ge 푑푣 + 푃∗ 푥 푣 = 푄∗ 푥 푑푥 d. Solve the linear equation.
Example
Solve the differential equation below
푑푦 푥 + 푦 = 푥2푦2 푑푥 Answer :
To solve the differential equations above, first we have to divided both side with 푥푦2. Here we get,
푑푦 푦−2 + 푥−1푦−1 = 푥 푑푥
푑푣 푑푣 푑푦 Suppose 푣 = 푦−1 then = −푦−2 or else = −푦−2 . By here, we get 푑푦 푑푥 푑푥
푑푣 + 푥−1푣 = 푥 푑푥 The differential equation above is linear equation. To solve the equation, we use integrating factor
1 푑푥 휇 푥 = 푒 푥 = 푒ln 푥 = 푥
Then the solution is
푣 = 휇−1 푥 휇 푥 푄 푥 푑푥 + 푐
1 1 푣 = 푥−1 푥2 푑푥 + 푐 → 푣 = 푥−1 푥3 + 푐 = 푥2 + 푐푥−1 3 3
1 푦−1 = 푥2 + 푐푥−1 3 1 1 = 푥2푦 + 푐푥−1푦 3
Exercise 11.
Solve the following differential equations.
푑푦 1 1. − 푦 = 푥푦2 푑푥 푥 푑푦 푦 2. + = 푦2 푑푥 푥 푑푦 1 3. + 푦 = 푒푥 푦4 푑푥 3 푑푦 4. 푥 + 푦 = 푥푦3 푑푥 푑푦 2 5. + 푦 = −푦2푥2 cos 푥 푑푥 푥
STATE UNIVERSITY OF-YOGYAKARTA FACULTY OF-MATHEMATICS AND NATURAL SCIENCE DEPARTMENT OF MATHEMATICS EDUCATION
Thirteenth II. First Order Differential Equations for Which Exact Solutions are obtainable Meeting : - Transformation
E. SPECIAL INTEGRATING FACTOR
For the last four meetings, we had solved four different differential equations which are not exact by transforming it into new equation which are exact. One of transformation that can be done is using integrating factor. Integrating factor can be used to transform non exact differential equation into exact differential equation. Unfortunately, there is no general expression for transformation. Nevertheless, we shall consider a few of transformation for special differential equations.
Consider the differential equation
푀 푥, 푦 푑푥 + 푁 푥, 푦 푑푦 = 0
Case Integrating factor 1 휕푀 푥, 푦 휕푁 푥, 푦 1 휕푀 푥,푦 휕푁 푥,푦 − − 푑푥 푁 푥, 푦 휕푦 휕푥 푒 푁 푥,푦 휕푦 휕푥 Depends upon x only
1 휕푁 푥, 푦 휕푀 푥, 푦 − 1 휕푁 푥,푦 휕푀 푥,푦 푀 푥, 푦 휕푥 휕푦 − 푑푦 Depends upon y only, then 푒 푀 푥,푦 휕푥 휕푦
Example
Solve it.
5푥푦 + 4푦2 + 1 푑푥 + 푥2 + 2푥푦 푑푦 = 0
Answer :
휕푀 푥, 푦 휕 5푥푦 + 4푦2 + 1 = = 5푥 + 8푦 휕푦 휕푦
휕푁 푥, 푦 휕 푥2 + 2푥푦 = = 2푥 + 2푦 휕푥 휕푥 1 휕푀 푥, 푦 휕푁 푥, 푦 1 − = 5푥 + 8푦 − 2푥 + 2푦 푁 푥, 푦 휕푦 휕푥 푥2 + 2푥푦
1 = 3푥 + 6푦 푥2 + 2푥푦
3 = 푥
1 휕푀 푥,푦 휕푁 푥,푦 Because − depends on x only, then the integrating factor of equation 푁 푥,푦 휕푦 휕푥 above is
1 휕푀 푥,푦 휕푁 푥,푦 3 − 푑푥 푑푥 푒 푁 푥,푦 휕푦 휕푥 = 푒 푥 = 푒3 ln 푥 = 푥3
Multiply both equation with 푥3, we get
푥3 5푥푦 + 4푦2 + 1 푑푥 + 푥3 푥2 + 2푥푦 푑푦 = 0
5푥4푦 + 4푥3푦2 + 1 푑푥 + 푥5 + 2푥4푦 푑푦 = 0
By using method of grouping as follow
5푥4푦 푑푥 + 푥5푑푦 + 4푥3푦2 푑푥 + 2푥4푦 푑푦 + 1 푑푥 = 0 and then define the total derivative as
푑 푥5푦 + 푑 푥4푦2 + 푑푥 = 0
Integrating it, we get the solution as follows
푥5푦 + 푥4푦2 + 푥 = 푐.
F. SPECIAL TRANSFORMATIONS
Consider the differential equation
푎1푥 + 푏1푦 + 푐1 푑푥 + 푎2푥 + 푏2푦 + 푐2 푑푦 = 0 where 푎1, 푏1, 푐1, 푎2, 푏2, and 푐2 are constants
Case Special Transformation 푎2 푏2 푥 = 푋 + 푕, 푦 = 푌 + 푘 ≠ Where (h,k) is the solution of the system 푎1 푏1 푎1푕 + 푏1푘 + 푐1 = 0 푎2푕 + 푏2푘 + 푐2 = 0 It will reduce into the homogenous differential equation 푎1푋 + 푏1푌 푑푋 + 푎2푋 + 푏2푌 푑푌 = 0 푎2 푏2 푧 = 푎1푥 + 푏1푦 = = 푘 will reduce the equation into separable 푎1 푏1 differential equation
Example
Solve this differential equation.
3푥 − 푦 + 1 푑푥 − 6푥 − 2푦 − 3 푑푦 = 0
Answer :
We can rewrite the equation above with
3푥 − 푦 + 1 푑푥 + −6푥 + 2푦 + 3 푑푦 = 0
For the equation above, 푎1 = 3, 푏1 = −1, 푎2 = −6 and 푏2 = 2. So, we get
푎 푏 2 = 2 = −2 푎1 푏1 Therefore, we can solve the equation by reduced it into separable differential equation using 푧 = 3푥 − 푦. This is implied that 푑푧 = 3푑푥 − 푑푦 or 푑푦 = 3푑푥 − 푑푧.
Substitute it into the question, we get
푧 + 1 푑푥 + −2푧 + 3 3푑푥 − 푑푧 = 0
푧 + 1 − 6푧 + 9 푑푥 + 2푧 − 3 푑푧 = 0
−5푧 + 10 푑푥 + 2푧 − 3 푑푧 = 0
2푧 − 3 푑푥 − 푑푧 = 0 5푧 − 10 By using some manipulations we get solution
2 1 푥 − 푧 + ln 10푧 − 20 = 푐 5 2
atau
2 1 푥 − 3푥 − 푦 + ln 10 3푥 − 푦. − 20 = 푐 5 2 Exercise 15.
Find the integrating factor and then solve the differential equations below. 1. 4푥푦 + 3푦2 − 푥 푑푥 + 푥 푥 + 2푦 푑푦 = 0 2. 푦 푥 + 푦 푑푥 + 푥 + 2푦 − 1 푑푦 = 0 3. 푦 푥 + 푦 + 1 푑푥 + 푥 푥 + 3푦 + 2 푑푦 = 0 4. 6푥 + 4푦 + 1 푑푥 + 4푥 + 2푦 + 2 푑푦 = 0 5. 3푥 − 푦 − 6 푑푥 + 푥 + 푦 + 2 푑푦 = 0 6. 2푥 + 3푦 + 1 푑푥 + 4푥 + 6푦 + 1 푑푦 = 0 7. 4푥 + 3푦 + 1 푑푥 + 푥 + 푦 + 1 푑푦 = 0
STATE UNIVERSITY OF-YOGYAKARTA FACULTY OF-MATHEMATICS AND NATURAL SCIENCE DEPARTMENT OF MATHEMATICS EDUCATION
Sixteenth Meeting III. Applications of First Order Equations : - Orthogonal Trajectories
As we had learned at the first meeting that differential equations were behind many mathematical models. In this chapter, we will discuss about the application of differential particularly for first order equations.
Before we study about the application of first order differential equation, first of all we will study about orthogonal trajectories and oblique trajectories.
Suppose it is given differential equation in derivative form
푑푦 = 푓(푥, 푦) 푑푥 The solution of its differential equation
퐹 푥, 푦 = 푐 or 퐹 푥, 푦, 푐 = 0
When we draw 퐹 푥, 푦, 푐 = 0, we can get a lot of curve depends on the value of c. By here, set of curves of 퐹 푥, 푦, 푐 = 0 is called one-parameter family of curves.
Definition
Let
퐹 푥, 푦, 푐 = 0
Be a given one-parameter family of curves in the XY plane. A curve that intersects the curves of the family at right angles is called an orthogonal trajectory of the given family.
Consider two families of curves F1 and F2. We say that F1 and F2 are orthogonal whenever any curve from F1 intersects any curve from F2, the two curves are orthogonal at the point of intersection.
Procedures to find the ortogonal trajectories of the family of curve, 퐹 푥, 푦, 푐 = 0
1. Find the differential equation of the given family, 푑푦 = 푓(푥, 푦) 푑푥 2. Find the differential equation of the orthogonal trajectories by replacing 푓 푥, 푦 by its negative reciprocal 1 − 푓(푥, 푦) 3. Solve the new differential equation. 푑푦 1 = − 푑푥 푓(푥, 푦)
Example
Find the orthogonal trajectories of the family of curve as follow :
푥2 + 푦2 = 퐶
Answer :
The differential equation related to the family of curve above is
2푥푑푥 + 2푦푑푦 = 0 or
푑푦 푥 = − 푑푥 푦
Therefore we get the differential equation of the orthogonal trajectories
푑푦 푦 = 푑푥 푥 1 1 푑푥 − 푑푦 = 0 푥 푦
Integrating both sides we get
ln 푥 − ln 푦 = ln 푐0 푥 = 푐 푦 Exercise 16.
1. 푦 = 퐶푒−2푥 2. 푥2 − 푦2 = 퐶 3. 푦 = 퐶푥2 4. 푥2 + 푦 − 푐 2 = 퐶2
STATE UNIVERSITY OF-YOGYAKARTA FACULTY OF-MATHEMATICS AND NATURAL SCIENCE DEPARTMENT OF MATHEMATICS EDUCATION
Seventeenth III. Applications of First Order Equations Meeting : - Oblique Trajectories
Definition
Let
퐹 푥, 푦, 푐 = 0 be a one-parameter family of curves. A curve that intersects the curves of this family at a constant angle 훼 ≠ 900 is called an oblique trajectory of the given family.
Procedures to find the oblique trajectories of the family of curve, 퐹 푥, 푦, 푐 = 0 at angle 훼 ≠ 900
1. Find the differential equation of the given family, 푑푦 = 푓(푥, 푦) 푑푥 2. Find the differential equation of the orthogonal trajectories by replacing 푓 푥, 푦 by expression 푓 푥, 푦 + tan 훼
1 − 푓 푥, 푦 tan 훼 3. Solve the new differential equation. 푑푦 푓 푥, 푦 + tan 훼 = 푑푥 1 − 푓 푥, 푦 tan 훼
Example
Find a family of oblique trajectories that intersect the family of straight lines 푦 = 푐푥 at angle 450.
Answer :
The derivative of the straight lines above is
푑푦 = 푐 푑푥 Therefore we get, the derivate of the oblique trajectory is
푑푦 푐 + tan 450 푐 + 1 = = 푑푥 1 − 푐 tan 450 1 − 푐 푐 + 1 푦 = 푥 1 − 푐 푐 + 1 or 푦 = 푑푥, where 푑 = 1 − 푐 Exercise 17.
1. Find a family of oblique trajectories that intersect the family of circles 푥2 + 푦2 = 푐 at angle 450. 2. Find a family of oblique trajectories that intersect the family of parabolas 푦2 = 푐푥 at angle 600. 3. Find a family of oblique trajectories that intersect the family of curves 푥 + 푦 = 푐푥2 at angle 훼 such that tan 훼 = 2.
STATE UNIVERSITY OF-YOGYAKARTA FACULTY OF-MATHEMATICS AND NATURAL SCIENCE DEPARTMENT OF MATHEMATICS EDUCATION
18th,19th Meeting : III. Applications of First Order Equations - Application in Mechanics Problems and Rate Problems
Next, it will be presented examples where differential equations are widely applied to model natural phenomena, engineering systems and many other situations.
III.A Exponential Growth – Population
Let 푃 푡 be a quantity that increases with time t. Suppose the rate of increase is proportional to the same quantity at that time. At first, the quantity is 푃0 > 0. This problem can be modeled into differential equation as follows
푑푃 = 푘푃, 푘 > 0 푑푡 where 푃 0 = 푃0.
This equation is called an exponential growth model.
The solution of its equation is
푘푡 푃 푡 = 푃0푒
III.B Exponential Decay
Let 푀 푡 be the amount of a product that decreases with time t. Suppose the rate of decrease is proportional to the amount of product at any time t. Suppose that at initial time
푡 = 0, the amount is 푀0. This problem can be modeled into differential equation as follows
푑푀 = −푘푀, 푘 > 0 푑푡
Where 푀 0 = 푀0.
This equation is called an exponential decay model.
The solution of its equation is
−푘푡 푀 푡 = 푀0푒
III.C Falling object
An object is dropped from a height at time 푡 = 0. If 푕 푡 is the height of the object at time t, 푎 푡 is the acceleration and 푣(푡) is the velocity, then the relationship between a, v and h are as follows :
푑푣 푑푕 푎 = , 푣 = 푑푡 푑푡 For a falling object, 푎 푡 is constant and is equal to 푔 = −9.8푚/푠.
Suppose that 푕0 and 푣0 are the initial height and initial velocity, respectively. The solution of the equations is
1 푕 푡 = 푔푡 + 푣 푡 + 푕 2 0 0 and is called the height of falling object at time t.
III.D Newton’s Law of Cooling
It is a model that describes, mathematically, the change in temperature of an object in a given environment. The law states that the rate of change (in time) of the temperature is proportional to the difference between the temperature T of the object and the temperature 푇푒 of the environment surrounding the object.
Assume that at t = 0, the temperature is 푇0.
This problem can be modeled as differential equation as follows
푑푇 = −푘(푇 − 푇 ) 푑푡 푒
Where 푇 0 = 푇0.
The final expression for 푇 푡 is given by
−푘푡 푇 푡 = 푇푒 + 푇0 − 푇푒 푒 This last expression shows how the temperature T of the object changes with time.
III.E RL Circuit
Let us consider the RL (resistor R and inductor L) circuit shown below. At 푡 = 0 the switch is closed and current passes through the circuit. Electricity laws state that the voltage across 푑푖 a resistor of resistance R is equal to R i and the voltage across an inductor L is given by 퐿 푑푡 (푖 is th current). Another law gives an equation relating all voltages in the above circuit as follows :
푑푖 퐿 + 푖푅 = 퐸 푑푡 Where E is a constant voltage and 푖 0 = 0.
Rewrite the differential equation above as
퐿 푑푖 = 1 퐸 − 푖푅 푑푡 퐿 −푅 − 푑푖 = 푑푡 푅 퐸 − 푖푅 Integrate both sides, we get
퐿 − ln 퐸 − 푖푅 = 푡 + 푐 푅 By using the initial condition, the formula of i is
푅푡 퐸 − 푖 = 1 − 푒 퐿 푅
Exercise 18, 19.
Find another application of first order linear differential equation.
STATE UNIVERSITY OF-YOGYAKARTA FACULTY OF-MATHEMATICS AND NATURAL SCIENCE DEPARTMENT OF MATHEMATICS EDUCATION
22nd, 23rd Meeting IV. Explicit Methods of Solving Higher Order Linear Differential Equations : - Basic Theory of Linear Differential Equations
The most general 푛푡푕 order linear differential equation is
푛 푛−1 ′ 푃푛 푥 푦 + 푃푛−1 푥 푦 + ⋯ + 푃1 푥 푦 + 푃표 푥 푦 = 퐺 푥 Where you’ll hopefully recall that
푑푚 푦 푦 푚 = 푑푥푚 It’s called homogeneous equation if 퐺 푥 = 0 and is called nonhomogeneous equation if 퐺 푥 ≠ 0.
Superposition Principle
Let 푦1, 푦2, … , 푦푚 be solutions of the homogeneous equation above. Then, the function
푦 = 푐1푦1 + 푐2푦2 + 푐3푦3 + ⋯ + 푐푛 푦푛 is also solution of the equation. This solution is called a linear combination of the functions
푦푖 .
The general solution of the homogeneous equation is given by
푦 = 푐1푦1 + 푐2푦2 + 푐3푦3 + ⋯ + 푐푛 푦푛
Where 푐1, 푐2, 푐3, … , 푐푛 are arbitrary constants and 푦1, 푦2, 푦3, … , 푦푛 are n-solutions of the homogeneous equation such that
푦1 푥 푦2 푥 푦3 푥 ⋯ 푦푛(푥) 푦′ 푥 푦′ 푥 푦′ 푥 푦′ (푥) 푊 푦 , 푦 , 푦 , … , 푦 푥 = 1 2 3 푛 ≠ 0 1 2 3 푛 ⋮ ⋮ ⋮ (푛−1) (푛−1) (푛−1) (푛−1) 푦 1 푥 푦 2 푥 푦 3 푥 ⋯ 푦 푛 푥 In this case, we will say that 푦1, 푦2, 푦3, … , 푦푛 are linearly independent. The function 푊 푦1, 푦2, 푦3, … , 푦푛 = 푊 is called the Wronskian of 푦1, 푦2, 푦3, … , 푦푛 .
The general solution of the nonhomogeneous equation is given by
푦 = 푐1푦1 + 푐2푦2 + 푐3푦3 + ⋯ + 푐푛푦푛 + 푦푝 = 푦푐 + 푦푝
Where 푐1, 푐2, 푐3, … , 푐푛 are arbitrary constants and 푦1, 푦2, 푦3, … , 푦푛 are linearly independent solutions of the associated homogeneous equation, and 푦푝 is a particular solution of nonhomogeneous equation.
Example.
Given that 푒푥 , 푒−푥 and 푒2푥 are the solutions of
푦 3 − 2푦" − 푦′ + 2푦 = 0
Show that these solutions are linearly independent on every real interval.
Answer :
푒푥 푒−푥 푒2푥 1 1 1 푊 푒푥 , 푒−푥, 푒2푥 = 푒푥 −푒−푥 2푒2푥 = 푒2푥 1 −1 2 = −6푒2푥 ≠ 0 푒푥 푒−푥 4푒2푥 1 1 4 Because the wronskrian is not equals to zero, then these solutions are linearly independent.
Exercise 22 1. Consider the differential equation 푑2푦 푑푦 푥2 − 2푥 + 2푦 = 0 푑푥2 푑푥 a. Show that 푥 and 푥2 are linearly independent solutions of this equation on the interval 0 < 푥 < ∞ b. Write the general solution of the given equation. c. Find the solution that satisfies the conditions 푦 1 = 3, 푦′ 1 = 2. Explain why this solution is unique. Over what interval is this solution defined?
2. Consider the differential equation 푑2푦 푑푦 푥2 + 푥 − 4푦 = 0 푑푥2 푑푥 a. Show that 푥2 and 푥−2 are linearly independent solutions of this equation on the interval 0 < 푥 < ∞. b. Write the general solution of the given equation. c. Find the solution that satisfies the condition 푦 2 = 3, 푦′ 2 = −1. Explain why this solution is unique. Over what interval is this solution defined? 3. Consider the differential equation 푦" − 5푦′ + 4푦 = 0 a. Show that each of the functions 푒푥 , 푒4푥and 2푒푥 − 3푒4푥 is a solution of this equation on the interval −∞ < 푥 < ∞. b. Show that 푒푥 and 푒4푥 are linearly independent on −∞ < 푥 < ∞. c. Show that the solution 푒푥 and 2푒푥 − 3푒4푥 are also linearly independent on −∞ < 푥 < ∞.
STATE UNIVERSITY OF-YOGYAKARTA FACULTY OF-MATHEMATICS AND NATURAL SCIENCE DEPARTMENT OF MATHEMATICS EDUCATION
24th, 25th, 26th IV. Explicit Methods of Solving Higher Order Linear Differential Equations Meeting : - The Homogeneous Linear Equation with Constant Coefficients
In this chapter we will be looking exclusively at linear second order differential equations. The most general linear second order differential equation is in the form
푛 푛−1 ′ 푃푛 푥 푦 + 푃푛−1 푥 푦 + ⋯ + 푃1 푥 푦 + 푃표 푥 푦 = 퐺 푥 In fact, we will rarely look at non-constant coefficient linear second order differential equations. In the case where we assume constant coefficients we will use the following differential equation.
푛 푛−1 ′ 푎푛 푦 + 푎푛−1푦 + ⋯ + 푎1푦 + 푎표 푦 = 퐺 푥 Initially we will make our life easier by looking at differential equations with 푔 푡 = 0.
we call the differential equation homogeneous and when 푔 푥 ≠ 0 we call the differential equation nonhomogeneous. So, let’s start thinking about how to go about solving a constant coefficient, homogeneous, linear, higher order differential equation. Here is the general constant coefficient, homogeneous, linear, higher order differential equation.
푛 푛−1 ′ 푎푛 푦 + 푎푛−1푦 + ⋯ + 푎1푦 + 푎표푦 = 0
All of the solutions in this example were in the form 푦 푥 = 푒푟푥 To see if we are correct all we need to do is plug this into the differential equation and see what happens. So, let’s get some derivatives and then plug in. 푦′ 푥 = 푟푒푟푥 , 푦 2 푡 = 푟2푒푟푥 , … , 푦 푛 푡 = 푟푛 푒푟푥 푛 푟푥 푛−1 푟푥 푟푥 푟푥 푎푛 푟 푒 + 푎푛−1 푟 푒 + ⋯ + 푎1푟푒 + 푎표 푒 = 0 푟푥 푛 푛−1 푒 푎푛 푟 + 푎푛−1푟 + ⋯ + 푎1푟 + 푎표 = 0
This can be reduced further by noting that exponentials are never zero. Therefore, the value of r should be the roots of 푛 푛−1 푎푛 푟 + 푎푛−1푟 + ⋯ + 푎1푟 + 푎표 = 0. This equation is typically called the characteristic equation for homogeneous equation.
Once we have these n roots, we have n solutions to the differential equation.
푟1푥 푟2푥 푟푛 푥 푦1 푥 = 푒 , 푦2 푥 = 푒 , … . . , 푦푛 푥 = 푒 You’ll notice that we neglected to mention whether or not the n solutions listed above are in fact “nice enough” to form the general solution. This was intentional. We have three cases that we need to look at and this will be addressed differently in each of these cases.
So, what are the cases? The roots will have three possible forms. These are
1. Real, distinct roots, .
2. Complex root,
3. Double roots, .
For repeated roots we just add in a t for each of the solutions past the first one until we have a total of k solutions. Again, we will leave it to you to compute the Wronskian to verify that these are in fact a set of linearly independent solutions. Finally we need to deal with complex roots. The biggest issue here is that we can now have repeated complex roots for 4th order or higher differential equations. We’ll start off by assuming that 푟 = 휆 ±
휇푖 occurs only once in the list of roots. In this case we’ll get the standard two solutions, 푒휆푥 cos 휇푥 푒휆푥 sin 휇푥
Example.
Solve the following equation.
푦 3 − 5푦" − 22푦′ + 56푦 = 0
Answer :
The characteristic equation is
푟3 − 5푟2 − 22푟 + 56 = 푟 + 4 푟 − 2 푟 − 7 = 0 Therefore we get
푟1 = −4, 푟2 = 2, 푟3 = 7 So we have three real distinct roots here and so the general solution is
−4푥 2푥 7푥 푦 푥 = 푐1푒 + 푐2푒 + 푐3푒
Example.
Solve the following differential equation
2푦 4 + 11푦 3 + 18푦" + 4푦′ − 8푦 = 0
Answer :
The characteristic equation is
2푟4 + 11푟3 + 18푟2 + 4푟 − 8 = 2푟 − 1 푟 + 2 3 = 0
1 So, we have two roots here, 푟 = and 푟 = −2 which is multiplicity of 3. Remember that 1 2 2 we’ll get three solutions for the second root and after the first we add x’s only the solution until we reach three solutions.
The general solution is then,
1 푥 −2푥 −2푥 2 −2푥 푦 푥 = 푐1푒2 + 푐2푒 + 푐3푥푒 + 푐4푥 푒
Example.
Solve the following differential equation.
푦 5 + 12푦 4 + 104푦 3 + 408푦" + 1156푦′ = 0
Answer:
The characteristic equation is
푟5 + 12푟4 + 104푟3 + 408푟2 + 1156푟 = 0
So, we have one real root r = 0 and a pair of complex roots 푟 = −3 ± 5 ± 푖 each with multiplicity 2. So, the solution for the real root is easy and for the complex roots we’ll get a total of 4 solutions. The general solution is
−3푥 −3푥 −3푥 −3푥 푦 푥 = 푐1 + 푐2푒 푐표푠 5푥 + 푐3푒 푠푖푛 5푥 + 푐4푥푒 푐표푠 5푥 + 푐5푥푒 푠푖푛 5푥
Exercise 24
Solve the problems below. 1. 푦" − 5푦′ + 6푦 = 0 2. 푦 3 − 3푦" − 푦 + 3푦 = 0 3. 푦" + 9푦 = 0 4. 푦 3 − 푦" + 푦 − 푦 = 0 5. 푦 4 + 8푦" + 16푦 = 0 6. 푦 5 − 2푦 4 + 푦 3 = 0 7. 푦 4 − 3푦 3 ′ − 2푦" + 2푦′ + 12푦 = 0
STATE UNIVERSITY OF-YOGYAKARTA FACULTY OF-MATHEMATICS AND NATURAL SCIENCE DEPARTMENT OF MATHEMATICS EDUCATION
27th, 28th Meeting IV. Explicit Methods of Solving Higher Order Linear Differential Equations : - The Method of Undetermined Coefficients
We now consider the n-order linear differential equation which is nonhomogeneous
푛 푛−1 ′ 푎푛 푦 + 푎푛−1푦 + ⋯ + 푎1푦 + 푎표 푦 = 퐺 푥 where 퐺 푥 ≠ 0. As we have studied before that the solution of nonhomogeneous equation above is in the form
푦 = 푦푐 + 푦푝
Where 푦푐 is the general solution for the corresponding homogeneous equation and 푦푝 is the particular solution. How to determine the particular solution, can be done with Undetermined Coefficient Method and Variation Parameters Method.
IV.A Undetermined Coefficient Method
If 퐺 푥 is an exponential function, polynomial, sine, cosine, sum/difference of one of these and/or a product of one of these then we guess the form of a particular solution. We then plug the guess into the differential equation, simplify and set the coefficients equal to solve for the constants. The one thing that we need to recall is that we first need the complimentary solution prior to making our guess for a particular solution. If any term in our guess is in the complimentary solution then we need to multiply the portion of our guess that contains that term by a x.
Example. Solve the following differential equation 푦 3 − 12푦" + 48푦′ − 64푦 = 12 − 32푒−8푥 + 2푒4푥 Answer : The characteristic equation corresponds to the homogeneous part is 푟3 − 12푟2 + 48푟 − 64 = 푟 − 4 3 = 0 So, the root is 4 of multiplicity 3. Therefore the complimentary solution is
4푥 4푥 2 4푥 푦푐 푥 = 푐1푒 + 푐2푥푒 + 푐3푥 푒 Based on the right side of the equation 12 − 32푒−8푥8푥 + 2푒4푥, our guess for a particular solution is
−8푥 4푥 푌푝 푥 = 퐴 + 퐵푒 8푥 + 퐶푒 Notice that the last term in our guess is the complimentary solution, so we’ll need to add one at least t to the third term in our guest. Also notice that multiplying in the third term by either 푥 or 푥2 will result in a new term that is still in the complimentary solution. So, we need to multiply the third term by 푥3 in order to get a term that is not contained in the complimentary solution. The final guess is then −8푥 3 4푥 푌푝 푥 = 퐴 + 퐵푒 8푥 + 퐶푥 푒 Substitute this function to the equation by taking it’s first, second and third derivative then set the coefficients equal we get, 3 1 1 퐴 = − , 퐵 = , 퐶 = 16 54 3 The general solution is then 3 1 1 푦 푥 = 푐 푒4푥 + 푐 푥푒4푥 + 푐 푥2푒4푥 − + 푒−8푥8푥 + 푥3푒4푥 1 2 3 16 54 3
Exercise 27 Find the general solution of the differential equations below
1. 푦" − 2푦′ − 8푦 = 4푒2푥 − 21푒−3푥 2. 푦" − 3푦 + 2푦 = 4푥2 3. 푦′′ + 2푦′ + 5푦 = 6 sin 2푥 + 7 cos 2푥 4. 푦′′ + 2푦′ + 2푦 = 10 sin 4푥 5. 푦′′ + 푦′ − 2푦 = 6푒−2푥 + 3푒푥 − 4푥2 6. 푦 3 − 4푦′′ + 5푦′ − 2푦 = 3푥2 푒푥 − 7푒푥
STATE UNIVERSITY OF-YOGYAKARTA FACULTY OF-MATHEMATICS AND NATURAL SCIENCE DEPARTMENT OF MATHEMATICS EDUCATION
29th, 30th Meeting IV. Explicit Methods of Solving Higher Order Linear Differential Equations : - Variation of Parameters
IV.B Variation of Parameters Methods
The method of variation of parameters involves trying to find a set of new functions,
푢1 푥 , 푢2 푥 , … , 푢푛 푥 so that,
푌 푥 = 푢1 푥 푦1 푥 + 푢2 푥 푦2 푥 + … + 푢푛 푥 will be a solution to the nonhomogeneous differential equation. In order to determine if this is possible, and to find the 푢푖 푥 if it’s possible, we’ll need a total of n equations involving the unknown functions that we can solve. Suppose
푦1 푥 푦2 푥 푦3 푥 ⋯ 푦푛(푥) 푦′ 푥 푦′ 푥 푦′ 푥 푦′ (푥) 푊 푦 , 푦 , … , 푦 푥 = 1 2 3 푛 1 2 푛 ⋮ ⋮ ⋮ (푛−1) (푛−1) (푛−1) (푛−1) 푦 1 푥 푦 2 푥 푦 3 푥 ⋯ 푦 푛 푥 th Let 푊푖 represent the determinant we get by replacing the i column of the Wronskian with the column 0,0,0, … ,0,1 . For example
0 푦2 푥 푦3 푥 ⋯ 푦푛 (푥) 0 푦′ 푥 푦′ 푥 푦′ (푥) 푊 = 2 3 푛 1 ⋮ ⋮ ⋮ (푛−1) (푛−1) (푛−1) 1 푦 2 푥 푦 3 푥 ⋯ 푦 푛 푥 the solution to the system can then be written as,
푔 푥 푊 푥 푔 푥 푊 푥 푔 푥 푊 푥 푢 푥 = 1 푑푥 , 푢 푥 = 2 푑푥 , … , 푢 푥 = 푛 푑푥 1 푊 푥 2 푊 푥 푛 푊 푥 Finally, the particular solution is given by
푔 푥 푊 푥 푔 푥 푊 푥 푔 푥 푊 푥 푌 푥 = 푦 (푥) 1 푑푥 + 푦 (푥) 2 푑푥 + … + 푦 (푥) 푛 푑푥 푝 1 푊 푥 2 푊 푥 푛 푊 푥
Example. Solve the following differential equation 푦 3 − 2푦" − 21푦′ − 18푦 = 3 + 4푒−푥
Answer : The characteristic equation corresponds to the homogeneous equation is 푟3 − 2푟2 − 21푟 − 18 = 0 → 푟 = −3, 푟 = −1, 푟 = 6 Here we get
−3푥 −푥 6푥 푦푐 푥 = 푐1푒 + 푐2푒 + 푐3푒 Several determinant that we should count are 푒−3푥 푒−푥 푒6푥 푊 = −3푒−3푥 −푒−푥 6푒6푥 = 126푒2푥 9푒−3푥 푒−푥 36푒6푥 0 푒−푥 푒6푥 −푥 6푥 5푥 푊1 = 0 −푒 6푒 = 7푒 1 푒−푥 36푒6푥 푒−3푥 0 푒6푥 −3푥 6푥 3푥 푊2 = −3푒 0 6푒 = −9푒 9푒−3푥 1 36푒6푥 푒−3푥 푒−푥 0 −3푥 −푥 −4푥 푊3 = −3푒 −푒 0 = 2푒 9푒−3푥 푒−푥 1 Here we get 3 + 4푒−푥 7푒5푥 1 푢 푥 = 푑푥 = 푒3푥 + 2푒2푥 1 126푒2푥 18 3 + 4푒−푥 −9푒3푥 1 푢 푥 = 푑푥 = − 3푒푥 + 4푥 2 126푒2푥 14 3 + 4푒−푥 2푒−4푥 1 1 4 푢 푥 = 푑푥 = − 푒−6푥 − 푒−7푥 3 126푒2푥 63 2 7 The general solution is then, 1 5 2 푦 푥 = 푐 푒−3푥 + 푐 푒−푥 + 푐 푒6푥 − + 푒−푥 − 푒−푥 1 2 3 6 49 7
Exercise 29 Solve the differential equations below. 1. 푦" − 2푦′ − 8푦 = 4푒2푥 − 21푒−3푥 2. 푦" − 3푦 + 2푦 = 4푥2 3. 푦′′ + 2푦′ + 5푦 = 6 sin 2푥 + 7 cos 2푥 4. 푦′′ + 2푦′ + 2푦 = 10 sin 4푥 5. 푦′′ + 푦′ − 2푦 = 6푒−2푥 + 3푒푥 − 4푥2 6. 푦 3 − 4푦′′ + 5푦′ − 2푦 = 3푥2 푒푥 − 7푒푥
STATE UNIVERSITY OF-YOGYAKARTA FACULTY OF-MATHEMATICS AND NATURAL SCIENCE DEPARTMENT OF MATHEMATICS EDUCATION
31st Meeting : EXERCISES
Question 1 : Solve the differential equation below. dy x 3y 5 dx x y 1 Question 2 : Sketch for the graph of solution of without finding it’s solution explicitly and give some reasons.
1 y (2 y)y , y(0) 2 Question 3 : Find the general solution of differential equation y 6y 9 0. Question 4 : Find the general solution of differential equation below. ty 2y sint ,t 0 Question 5 : Find the general solution of the differential equation below. (x2 y)dx (x e y )dy 0 Question 6 : Find the general solution of the second order linear differential equation below. y 2y 3 4sin 2t