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Department of Engineering Mathematics EMAT10100 Eng Maths 1 - Calculus Lecture 12: Higher Order ODEs - Particular Solutions In the previous lecture we studied solutions of second order, linear, homogeneous ODEs with constant coeﬃcients. Recall that equations of this type: x ˙ a¨ + bx + cx = 0, can be solved by computing roots of: am2 + bm + c = 0, resulting in 2 roots, m1 and m2 . This results in 3 scenarios: 1. m1 = m2 ∈ R resulting in the general solution: x(t) = Cem1 t + Dem2 t 2. m1 = m2 = m ∈ R resulting in the general solution: x(t) = Cemt + Dtemt 3. m1,2 = α ± jβ - a complex conjugate pair, results in the general solution: x(t) = eαt [C cos (βt) + D sin(βt)] In this ﬁnal lecture, we consider how to solve non-homogenous equations. We again consider only those with constant coeﬃcients, i.e. of the form: d2 x dx a 2 +b + cx = f(t), dt dt where f(t) = 0. Noting that the ODE of interest may be rewritten as the LINEAR OPERATOR L: L[x] = f(t), then it can be shown that the most general solution of the ODE is: x(t) = x(t) + xp (t), where • x(t) is the solution of the associated homogeneous equation: L[x] = 0, which is termed the COMPLEMENTARY FUNCTION. • xp (t) is any solution you can ﬁnd of the full non-homogenous equation, which is termed the PARTICULAR SOLUTION. This result is easy to prove. (i) x(t) is a solution of L[x] = 0 (ii) xp (t) is a solution of L[x] = f(t) then: L[x + xp ] = f(t) From (i) we have L[x] = 0 and from (ii) L[xp ] = f(t) then: L[x + xp ] = L[x] + L[xp ] = f(t). Unfortunately, that is as good as it gets as currently we have no general methodology for deter- mining the particular solution xp (t) for any given f(t) and we must resort to essentially trial and error (with some educated guessing thrown in for good measure!) Example: Consider the equation: x + x − 2x = 3e−t ¨ ˙ The solution is given by x(t) = x(t) + xp (t) To ﬁnd x we solve the associated homogeneous equation: ¨ ˙ x + x − 2x = 0 ⇒ m2 + m − 2 = 0 ⇒ m1 = −2 and m2 = 1 From this we can conclude that the complementary function is given by: x(t) = Ae−2t + Bet Now to ﬁnd the particular solution. Since f(t) = 3e−t , our trial function must be something that when diﬀerentiated involves e−t , so we try a trial function xp (t) = Ce−t . To check whether this trial function is a solution of the DE, we substitute xp (t) = Ce−t into the DE by observing that: xp = −Ce−t ˙ xp = Ce−t ¨ we substitute these values as follows: ¨ xp ˙ +xp −2xp = 3e−t , ⇒ Ce −t −Ce −t −2Ce−t = 3e−t , ⇔ −2C = 3 3 ⇒ C=− . 2 Therefore: 3 xp (t) = − e−t 2 From these we can write down the GENERAL SOLUTION of the ODE: 3 x(t) = x + xp = Ae−2t + Bet − e−t 2 Note that you should only make use of any initial conditions once you have written down the general solution. If you attempt to ﬁnd the parameters A and B before you have evaluated the particular solution, you will not ﬁnd the correct parameter values! Example: (The Failure Case) Suppose that we now consider a similar looking ODE: x − 5x + 4x = et ¨ ˙ We ﬁrst ﬁnd the complementary function by evaluating the roots of the characteristic polynomial associated with the ODE: m2 − 5m + 4 = 0 ⇒ m1 = 4, m2 = 1. Therefore the complementary function is given by: x(t) = Ae4t + Bet . Now to ﬁnd the particular solution, we would be tempted to try a trial function of the form: xp (t) = Ce−t , as before since the forcing term involves the function et . However, e−t forms part of the complementary function, which we know when diﬀerentiated will give a solution to the homogeneous equation, which is inconsistent with the solution we need. What should we do instead? You may recall from our study of the homogeneous case, that when we had repeated roots of the equation, we consider a solution that involved multiplying the root by a power of t. We shall try the same trick again and assume that our trial solution takes the form xp (t) = Cte−t instead. In this case: xp = Ctet ⇒ x˙p = C(tet + et ), ⇒ xp = C(tet + 2et ), ¨ so that C(tet + 2et ) − 5C(tet + et ) + 4Ctet = et 1 i.e. C = − 3 . Thus the General Solution in this case is given by: 1 x(t) = Ae4 t + Bet − tet . 3 Table 1: In general, the choice of xp (t) can be done using the following guidelines Case f(t) xp (t) kt (i) Ae αekt (ii) f(t) is an exponential (ekt ) which features in the CF αtekt A cos(ωt) or (iii) C sin(ωt) + D cos(ωt) A sin(ωt) (iv) f(t) is a polynomial of order n polynomial in t of same order As we mentioned before, ﬁnding particular solutions is in general not possible. Essentially if we do not have a RHS that is of the above form then ﬁnding the particular solution isn’t going to happen! Example Find the particular solution of: ¨ ˙ x − 5x + 6x = 24t In this case we consider a particular solution of the form: xp (t) = K1 t + K2 ¨ In this case we see that x˙p = K1 and xp = 0. Substituting these into the ODE we have obtain: −5K1 + 6(K1 t + K2 ) = 24t From this we have two simultaneous equations: 6K1 = 24, −5K1 + 6K2 = 0. 10 This tells us that K1 = 4 and K2 = 3 . Example Solve the ODE d2 y dy 2 +3 + 2y = 10 cos x, y(0) = 1, y (0) = 0 dx dx This problem is slightly diﬀerent in that we have some initial data, that we will utilise to evaluate the coeﬃcients of the complementary function. The ﬁrst step though is to solve the equation and we proceed in a similar manner to before. We ﬁrst ﬁnd the complementary function by evaluating the roots of the characteristic polynomial: m2 + 3m + 2 = 0 ⇒ m1 = −2, m1 = −1, from which we write down the CF: y = Ae−x + Be−2x . Consulting table 1, we shall try a trial solution of the form: yp (x) = λ cos x + µ sin x, ⇒ yp (x) = −λ sin x + µ cos x, ⇒ yp (x) = −λ cos x − µ sin x. substituting these into the ODE we obtain: (−λ cos x − µ sin x) + 3(−λ sin x + µ cos x) + 2(λ cos x + µ sin x) = 10 cos x. Equating coeﬃcients of cos x and sin x we see that: λ + 3µ = 10 ⇒ λ = 1, µ=3 3λ − µ = 0 This gives us the general solution: y(x) = Ae−x + Be−2x + cos x + 3 sin x. To ﬁnd the values of the parameters A and B, we make use of the initial data: y(0) = 1, y (0) = 0, which gives us another set of simultaneous equations to solve for A and B: 1=A+B+1 ⇒ A = −3, B = 3. 0 = −A − 2B + 3 This results in the answer: y(x) = 3e−2x − 3e−x + cos x + 3 sin x. Exercises: 1. Suppose that: ¨ x + x = t, and that x(0) = 0 and x( π ) = 0. Determine the value of x when t = π . 2 4 2. Evaluate the general solution of the ODE: d2 y dy 2 +8 + 25y = 48 cos x − 16 sin x. dx dx Find also the particular solution given y(0) = 8 and y (0) = −27. 3. Find the particular solution of the ODE: x − 5x + 6x = (12t − 7)e−t ¨ ˙ HINT: You should try a trial solution, xp (t) of the form (At + B)e−t . HOMEWORK: Read James pg. 723–734. Exercises 10.9.2 and 10.9.4 Well that about wraps it up for this part of the course. In the next part of the course we will be considering probability theory. John R. Terry Weeks 11-16

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Lecture 12 Higher Order ODEs - Particular Solutions

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