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MATH201 (10)
Chapter 3

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Department
Mathematics
Course
MATH201
Professor
Callum Quigley
Semester
Winter

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Math 201 3.4. VARIABLE COEFFICIENT EQUATIONS 31 2 ′′ ′ Example 3.29. Consider the equation x y + xy − y = 0. By inspection, ϕ (x) = x is a solu1ion. Let ϕ 2x) = v(x)ϕ (x1 = xv(x). Then ϕ = xv + v, ϕ = xv + 2v . ′ 2 2 The equation becomes xv + 3v = 0. Letting w = v we get ′ dw 3 1 1 1 = − dx =⇒ w(x) = 3 =⇒ v(x) = − 2 =⇒ ϕ2(x) = − . w x x x x Notice that ϕ (x) 1 2 = − ▯= constant ϕ1(x) x2 c2 so that ϕ 1nd ϕ are2linearly independent. The general solution to the original de is y(x) = c x + 1 . x Another concept that will prove to be useful in characterizing solutions to linear diﬀerential equations is that of the Wronskian. Deﬁnition 3.30. The Wronskian of two functions f and g is deﬁned as f(x) g(x) W[f,g](x) := ′ ′ . ◭ f (x) g (x) Lemma 3.31. If (i) P and Q are continuous on (a,b); and (ii) the functions ϕ and ϕ are solu1ions to 2 ′′ ′ y + P(x)y + Q(x)y = 0 on (a,b), then either W[ϕ ,ϕ ] is i1ent2cally zero on (a,b); or W[ϕ ,ϕ ] is never1 2 zero on (a,b). Proof . Diﬀerentiate the Wronskian ϕ1 ϕ2 ′ ′ W[ϕ ,1 ]2= ′ ′ = ϕ 1 −2ϕ ϕ 1 2 ϕ1 ϕ2 to get dW ′′ ′ ′ ′ ′ ′′ ′′ ′′ = ϕ 1 +2ϕ ϕ −1ϕ 2 − ϕ 1 =2ϕ ϕ 1 ϕ2ϕ 1 2 1 2 dx = ϕ 1−P(x)ϕ − Q(x)ϕ ) − ϕ (2P(x)ϕ 2 Q(x)ϕ ) ′ 1 2 1 = −P(x)(ϕ ϕ 1 ′ − ϕ ′ϕ 2 − Q(x)(ϕ ϕ 1 ϕ2ϕ ) =1−P2x)W. 2 1 R Hence W = Ae − P(x) dx. If A = 0, then W ≡ 0, and if A ▯= 0, then W = ▯ 0 for all x ∈ (a,b). The following result gives a nice characterizationof the solutions of linear, homogeneousdiﬀerential equations. Theorem 3.32. If (i) P and Q are continuous on (a,b); and (ii) ϕ1and ϕ ar2 solutions to y + P(x)y + Q(x)y = 0 on (a,b), then ϕ 1nd ϕ are2linearly independent iﬀ W[ϕ ,ϕ ] ▯= 1 fo2 all x ∈ (a,b). Math 201 3.5. VARIATION OF PARAMETERS 32 3.5 Variation of Parameters Consider the nonhomogeneous equation y + P(x)y + Q(x)y = g(x). (3.12) In a previous section the method of undetermined coeﬃcients was used to ﬁnd particular solutions to equa- tions for which the linear operator had constant coeﬃcients. Even then, it only works for a certain type of nonhomogeneous term, one for which the form of the particular solution can be guessed. In this section we discuss a method more general that the method of undetermined coeﬃcients. The method called variation of parameters, that we will discuss here, has one major advantage over the method of undetermined coeﬃcients. It always works, even for general coeﬃcients P(x) and Q(x), and it requires no guessing, provided that two linearly independent solutions to the homogeneous equation are known. So why did we bother with the method of undetermined coeﬃcients? Because the method of variation of parameters can sometimes be diﬃcult or tedious to implement, and if one can guess the form of a particular solution, it is sometimes easier to use the method of undetermined coeﬃcients. We now derive the procedure known as variation of parameters. First consider the homogeneous counterpart to (3.12), namely y + P(x)y + Q(x)y = 0. (3.13) Suppose that {ϕ (x),1 (x)} a2e linearly independent solutions to (3.13), so that the general solution to (3.13) is y(x) = c 1 (1) + c ϕ 2x)2 The idea behind this method is to look for a particular solution to (3.12) of the form ϕ px) = v (1)ϕ (x)1+ v (x)ϕ2(x). 2 (3.14) This is the homogeneous solution in which the constants, or parameters, c and c are a1lowed to2vary, hence the name variation of parameters. By looking for a particular solution of this form, we have introduced two new unknown functions, v and v 1 so we 2ill require two conditions to solve for them. One condition will, of course, come from requiring that ϕ satisfyp(3.12). But what about a second condition? As there is no obvious condition to impose, we will choose a condition to impose that is convenient. Diﬀerentiating Eq. (3.14) we get ′ ′ ′ ′ ′ ϕ px) = v (x1ϕ (x)1+ v (x)ϕ2(x) + 2 (x)ϕ (x1 + v (1)ϕ (x). 2 2 Since we need a second condition, we simplify the above expression by imposing the following condition: ′ ′ v1(x)ϕ 1x) + v (x2ϕ (x)2= 0. (3.15) ′ Then the previous expression for ϕ reducep to ′ ′ ′ ϕ px) = v (1)ϕ (x)1+ v (x)ϕ2(x). 2 Diﬀerentiating again, we get ′′ ′ ′ ′ ′ ′′ ′′ ϕ p = v 1 1 + v 2 2+ v 1 1 + v 2 2 . Insert these quantities into (3.12) to get v1ϕ 1 v ϕ 2 v2ϕ + v1ϕ 1 P(x)2v2ϕ + v ϕ ) + 1(x1(v ϕ 2 v2ϕ ) = g(x). 1 1 2 2 Math 201 3.5. VARIATION OF PARAMETERS 33 Re-arranging this equation we get v ϕ + v ϕ + v [ϕ + P(x)ϕ + Q(x)ϕ ] + v [ϕ + P(x)ϕ + Q(x)ϕ ] = g(x). 1 1 2 2 1 1 1 1 2 2 2 2 Notice that the terms in square brackets are zero since ϕ and ϕ are solutions to the homogeneous equation 1 2 (3.13), so that the above equation reduces to ′ ′ ′ ′ v1ϕ 1 v ϕ2= 2(x). (3.16) ′ ′ Equations (3.15) and (3.16) form a pair of equations for the two unknown functions v and v , wh1ch can 2e written in matrix form as follows: ϕ ϕ v ′ 0 1 2 1 = . ϕ 1 ϕ 2 v2′ g This matrix equation has a solution provided its coeﬃcient matrix is non-singular. But this is in fact true since the determinant of the coeﬃcient matrix is just the Wronskian W[ϕ ,ϕ ], which1by 2heorem 3.32, is non-zero for linearly independent solutions of the homogeneous equation (3.13). Thus, this equation can be solved by Cramer’s rule yielding 0 ϕ2 ϕ1 0 ′ ′ ′ g ϕ2 ′ ϕ1 g v1= , v2 = , ϕ 1 ϕ 2 ϕ 1 ϕ 2 ϕ ′ ϕ ′ ϕ ′ ϕ ′ 1 2 1 2 which can be written as ϕ (x)g(x) ϕ (x)g(x) v (x) = − 2 , v′(x) = 1 . 1 W[ϕ ,1 ](2) 2 W[ϕ ,1 ](2) Thus, v 1nd v are2obtained by integration, and the particular solution (3.14) becomes Z x ϕ1(ξ)ϕ (2) − ϕ (x1ϕ (ξ)2 ϕp(x) = g(ξ)dξ. x0 W[ϕ ,1 ](2) ′′ Example 3.33. Find the general solution to the equation y + y = tanx on the interval (−π/2,π/2). Solution ′′ The homogeneous equation y + y = 0 has two linearly independent solutions given by ϕ1(x) = cosx, ϕ 2x) = sinx. The Wronskian for these is cosx sinx 2 2 W[ϕ ,1 ] 2 = cos x + sin x = 1. −sinx cosx We look for a particular solution of the form ϕ (x) = v (x)ϕ (x) + v (x)ϕ (x) = v (x)cosx + v (x)sinx. p 1 1 2 2 1 2 The functions v an1 v sati2fy ′ ϕ2(x)g(x) ′ ϕ 1x)g(x) v1(x) = − = −sin x tanx, v2(x) = = cosx tanx. W[ϕ ,ϕ1](2) W[ϕ ,1 ](2) Math 201 3.6. APPLICATIONS 34 Integration yields Z Z Z sin x cos x − 1 v1(x) = − sinx tanxdx = − dx = dx Z cosx cosx = (cosx − secx)dx = sinx − ln(secx + tanx) + c 1 and Z Z v2(x) = co
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