EUT CHAPTER 2 : First Order Differential Equations

EUT 1022 Learning Outcomes a) At the end of this chapter, it is expected that all students will be able to know: What is ODE? What is ODE? What is order of ODE? What is order of ODE? How to solve ODE? How to solve ODE? How to use IVP and BVP? How to use IVP and BVP? b) And also at the end of the chapter students will have to possess the skills of solving first-order ODE.

EUT An equation that contains an unknown function and one or more of its derivatives WHAT IS A DIFFERENTIAL EQUATION?

EUT ODE BASICS “An ordinary differential equation (ODE) is an equation that contains one or several derivatives of an unknown function.” (Kreyzig). For example:

EUT Order of a Differential Equation The order of a differential equation is the order of the highest derivative that appears in the equation. First order differential equations can be written as:

EUT Example 2.1

EUT Solution of an ODE A function f is called a solution of a differential equation if the equation is satisfied when y=f(x) and its derivatives are substituted into the equation.

EUT Example Solve the following differential equations :

EUT FIRST ORDER DIFFERENTIAL EQUATIONS Separable Equations Homogeneous Equations Linear Equations Exact Equations Applications

EUT : SEPARABLE EQUATIONS The differential equation ; y’ = f(x,y) is said to be separable if the equation can be written as the product of a function of x, u(x) and a function of y,v(y) i.e. y’ = u(x)v(y)

EUT Example 2.2 y’ - xy = x can be written as y’ = x(1+y) i.e u(x) = x & v(y) = 1+y y’ siny cosx – cosy sinx = 0 or can be written as y’ = tan x cot y

EUT Exercise Determine each of the following ODEs is separable equations or not. a) y’ = xe y-x b) y’x = x – 2y

EUT Solution of Separable Equations The equation y’ = u(x)v(y) can be written in the form ; Then, we integrate on both sides of the equation: where A is constant

EUT Example 2.3 Solve the following differential equations : a)(x + 2)y’ = y b) y’e x + xy 2 = 0 c)x 2 y’ = 1 + y

EUT Solutions a) y -1 dy = (x+2) -1 dx By integrating both sides : ln |y| = ln |x+2| + A ln |y| - ln |x+2| = A and becomes y = B(x+2).

EUT Example 2.4 Let θ be the temperature (in o C) for a mass in a room with a constant room temperature at 18 o C. The mass cools from 70 o C to 57 o C in 5 minutes, how much longer will be needed by the mass to cool down to 40 o C.

EUT Solution Hint…

EUT = 18 + Ae -kt – From the initial condition, we obtain A = 52 and k = – Solve … The mass will need approximately minutes more to cool.

EUT : HOMOGENEOUS EQUATIONS Definition 2.2 : A differential equation y’ = f(x,y) is said to be homogeneous if f(λx, λy) = f(x, y) for any real number λ.

EUT Example 2.5 Determine whether y’y = x(lny – lnx) is a homogeneous equation.

EUT Solution Then, show that f(λx, λy) = f(x,y).

EUT Example 2.6

EUT Solution of Homogeneous Equations A homogeneous equation can be transformed into separable equations by substituting y = xv or v = y/x. Hence, y’ = v + x (dv/dx) This will result in separable equations in variables x and v. Solve and resubstitute the value v=y/x.

EUT Example 2.7 Solve the initial value problem. y(0) = 2.

EUT Solution Test the homogenity Substitute y = xv Integrate

EUT Substitute v = y/x and determine the general solution. The initial condition, y(0) = 2, so that A = 2.

EUT Sekalipun mereka dapat memahami hakikat ‘bagaimana hendak bermula tetapi tidak pernah bersiap’…. Sampai bila pun mereka tidak akan bermula “Peluang sentiasa ada kepada setiap orang. Hanya kesanggupan yang membezakan”

EUT Example 2.8

EUT

EUT : LINEAR EQUATIONS Definition 2.3 : (First OLDEs) An equation is said to be a first-order linear equation if it has the form a(x)y’ + b(x)y = c(x) where a(x), b(x) & c(x) are continuous functions on a given interval.(or a constant).

EUT Example 2.9 a) xy’ – 2y = x + 1 is a linear equation with a(x) = x, b(x) = -2 dan c(x) = x+1 b) 2x 2 y’ + xe y = sin x is not a linear equation.

EUT Example 2.10 Determine whether the following ODEs are linear equations or not. a) (1 - x 2 )y’ = x(y + sin -1 x) b) y’ + xy 2 = e x

EUT Solution of Linear Equations We rewrite the linear the equation as y’ + p(x)y = q(x) a(x) = 1 find ∫p(x)dx An integrating factor is ρ = e ∫p(x)dx

EUT Cont..... multiplying both sides of the linear equation by ρ we get, ρy ’ + ρp(x)y = ρq(x) Or can be rewrite into the form; The solution to the linear equation is by integrating both sides with respect to x.

EUT Example 2.11 Solve the following linear equations. a) y’ + y = x b) y’ +y tan x = kos x, given that y = 1 when x = 0.

EUT Solutions a) y’ + y = x we know that p(x) = 1 and q(x) = x. Answer : y = (x – 1) + Ae -x b) Answer : y = (x + 1) kos x

EUT Example 2.12: Linear Equations Application A tank containing 50 liters of liquid with composition, 90% water and 10% alcohol. A second liquid with composition, 50% water and 50% alcohol is poured into the tank at the rate of 4 liters per minute. While the second liquid is poured into the tank, the liquid in the tank is drained out at the rate of 5 liters per minute. Assuming that the liquid in the tank mix uniformly, how much alcohol left in the tank 10 minutes later?

EUT Answer liter

EUT : EXACT EQUATIONS Definition 2.4 A first order differential equation of the form M(x,y)dx + N(x,y)dy = 0 is called an exact diffrential equation if the differential form M(x,y)dx + N(x,y)dy is exact,that is, du = M(x,y)dx + N(x,y)dy of some function u(x,y).

EUT Theorem 2.1 : (Condition of an exact equation) * The equation M(x,y)dx + N(x,y)dy = 0 is an exact equation if and only if :

EUT Example 2.13 Show that (6xy+2y)y’ = –(2x + 3y 2 ) is an exact equation.

EUT Solution We rewrite in the form (2x + 3y 2 )dx + (6xy + 2y)dy = 0 i.e. M = (2x + 3y 2 ) and N = (6xy + 2y) Then, we test for exactness

EUT Example 2.14 Solve the following differential equation; (1- kos 2 x) dy + (y sin 2x) dx = 0. Solution :

EUT Solution of An Exact Equations 1) Write in the form of exact equation : M(x,y)dx + N(x,y)dy = 0. Test for Exactness :

EUT Cont… 2) Write

EUT Cont.... 3) By integrating Equation (I) with respect to x, we have u(x,y) = ∫ M dx + Φ(y). 4) Differentiate u with respect to y, and equating the result with equation (II) to determine the function Φ(y). 5) Write the solution in the form of u(x,y) = A, where A is a constant. 5) By substituting the initial condition, we will get the particular solution of the initial value problem

EUT Example 2.14 Show the the following ODE is an exact equation and solve the differential equation. (6x xy + 3y 2 )dx + (-5x 2 + 6xy -3y 2 )dy = 0

EUT Solution Test for Exactness (6x xy + 3y 2 ) = M and (-5x 2 + 6xy -3y 2 ) = N Show that

EUT Cont … Let u(x,y) be the solution, then u = ∫Mdx +Φ(y) u = ∫(6x xy + 3y 2 )dx + Φ(y) = 2x 3 – 5x 2 y +3xy 2 + Φ(y) differentiate u with respect to y, we obtain ∂u/∂y = -5x 2 +6xy + Φ ’ (y)

EUT Cont..... By equating with N we obtain, -5x 2 +6xy + Φ ’ (y) = (-5x 2 + 6xy -3y 2 ) hence; Φ ’ (y) = -3y 2 and we obtain Φ(y) = -y 3 + B The general solution is u(x,y) = A, where u(x,y) = 2x 3 – 5x 2 y +3xy 2 - y 3 + B = A u(x,y) = 2x 3 – 5x 2 y +3xy 2 - y 3 = C, C = A - B

EUT Example 2.15 Solve the following differential equations 1) xy’ + y + 4 = 0 2) sin x dy + (y kos x – x sin x) dx = 0

EUT Answer 1) u(x,y) = (y + 4)x = C, C = A – B 2) u(x,y) = y sin x + x kos x – sin x = C, C = A – B

EUT Exercise 2 Show that the following differential equation 2y dx + x dy = 0 is not an exact equation. If the integrating factor, solve the differential equation.

EUT Answer x 2 y = A, A is a constant