Presentation on theme: "First Order Linear Differential Equations"— Presentation transcript:
1 First Order Linear Differential Equations Any equation containing a derivative is called a differential equation.A function which satisfies the equation is called a solution to the differential equation.The order of the differential equation is the order of the highest derivative involved.The degree of the differential equation is the degree of the power of the highest derivative involved.We can then integrate both sides.This will obtain the general solution.
2 A first order linear differential equation is an equation of the form 1To find a method for solving this equation, lets consider the simpler equationWhich can be solved by separating the variables.(Remember: a solution is of the form y = f(x) )
3 orUsing the product rule to differentiate the LHS we get:
4 Returning to equation 1,If we multiply both sides byNow integrate both sides.For this to work we need to be able to find
8 To solve this integration we need to use substitution. If you look at page 113 you will see a note regarding the constant when finding the integrating factor. We need not find it as it cancels out when we multiply both sides of the equation.
9 (note the shortcut I have taken here) The modulus vanishes as we will have either both positive on either side or both negative. Their effect is cancelled.(note the shortcut I have taken here)
10 (note the shortcut I have taken here) The modulus vanishes as we will have either both positive on either side or both negative. Their effect is cancelled.(note the shortcut I have taken here)
11 So far we have looked at the general solutions only So far we have looked at the general solutions only. They represent a whole family of curves. If one point can be found which lies on the desired curve, then a unique solution can be identified.Such a point is often given and its coordinates are referred to as the initial conditions or values. The unique solution is called a particular solution.
16 The solutions to this quadratic will provide two values of m which will make y = Aemx a solution.
17 If we call these two values m1 and m2, then we have two solutions. A and B are used to distinguish the two arbitrary constants.Is a solution.The two arbitrary constants needed for second order differential equations ensure all solutions are covered.The type of solution we get depends on the nature of the roots of this equation.
18 When roots are real and distinct The auxiliary equation isTo find a particular solution we must be given enough information.
19 The auxiliary equation is Using the initial conditions.
20 Solving givesThus the particular solution isPage 119 Exercise 3 Questions 1(a), (b) 2(a), (b)But we will now look at the solution when the roots are real and coincident.
21 Roots are real and coincident When the roots of the auxiliary equation are both real and equal to m, then the solution would appear to be y = Aemx + Bemx = (A+B)emxA + B however is equivalent to a single constant and second order equations need two.With a little further searching we find that y = Bxemx is a solution. Proof on p119So a general solution is
29 Non homogeneous second order differential equations Non homogeneous equations take the formSuppose g(x) is a particular solution to this equation. ThenNow suppose that g(x) + k(x) is another solution. Then
30 GivingFrom the work in previous exercises we know how to find k(x).This function is referred to as the Complimentary Function. (CF)The function g(x) is referred to as the Particular Integral. (PI)General Solution = CF + PI
31 Finding the (CF): the auxiliary equation is Substituting into the original equation
35 Finding the form of the PI The individual terms of the CF make the LHS zeroThe PI makes the LHS equal to Q(x). Since Q(x) ≠ 0, it stands to reason that the PI cannot have the same form as the CF.When choosing the form of the PI, we usually select the same form as Q(x).This reasoning leads us to select the PI according to the following steps.try the same form as Q(x)If this is the same form as a term of the CF, then try xQ(x)If this is the same form as a term of the CF, then try x2Q(x)
37 Finding the (CF): the auxiliary equation is Substituting into the original equation
38 If the wrong selection of PI is made, you will generally be alerted to this by the occurrence of some contradiction in later work.Page 126 – Exercise 7A as many as possible.TJ Exercise 3. This assesses beyond poly and trig work.