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# diffeqs - 9.1 First Order Linear Equations A differential...

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Unformatted text preview: 9.1 First Order Linear Equations A differential equation is an equation such as ' ® — z 3 1 w y () Where y is a function of :13. That is, it involves derivatives of an unknown function (in this case 31(3)). Differential equations are used in physics, engineering... to describe physical processes, and usually we want to try solve for y (x) You can check that if C is a constant, y (m) = 063“: satisﬁes (1). It is called a solution of (1). The order of the differential equation is the order of the highest derivative appearing in the equation. The equation above is ﬁrst order, since 3% is a ﬁrst derivative. Similarly, 2 g + 3% dm2 aim is a second order equation, since the highest order derivative is the second +y=0 derivative g? In this and the next section, we shall- show howr to solve two special types of differential equations. First Order Linear Equations A differential equation of the form r+pew=qe) ' (a where p (m) and q (3:) are given functions of m, is called a ﬁrst order linear equation. The functions p (m) and (1(5):) are assumed to be continuous on the interval -I on which we want to solve for y (x) There is a fairly straightforward algorithm to solve for y : Method of Solution Step 1 Find the integrating factor Calculate H(m) = /p(m)d\$ (without a constant of integration) and then form 8H”). The function 8H is called the integrating factor. 85 Step 2 Multiply by the integrating factor We multiply the differential equation (2) by BH : eH(m)y' (a?) + eH(‘”)p (3:) y (as) = equ (cc). Then we claim that we can rewrite this as dig: (eHWy (a) = eHWq (m). Proof of the claim By the chain rule, d H(a:) _ Hmdﬂ _ Hm 0393(3 )—e (ix—e 33(33) So, d HCm) e d Hm Heady dm<e y(\$)) _ dm (8 ) Mm)” "Ci—.45 = 6mm}? (3) CU (=1?) + Emmy, (13) a which is the same as the left—hand side in (3) I Step 3 Integrate (4) We integrate (4), namely d d—m- (emﬂy (33)) : 8H(“’)Q(w), giving (23% (so) = f amass) are + c, where C is a constant of integration. Then the solution is y {:6} = e—H(m) U eHWq (comm + a] . Remarks (3) (4) (a) Note that the solution involves an unknown constant of integration. Some— times we use other information to solve for C. (b) For a given equation, you will often have to recognize what are the functions p (3:) and q (51:). Moreover, you may have to manipulate it into the form y’+p(m)y:q(w), if it is not already in that form. 86 Example 1 Let (2,1) be real numbers with a % O. Solve the equation y' + ay 2 b. (1) Solution We see that the equation has the form y’ +P(I)y = Mm) where Mm) = a;q(\$) = b are constant functions. Step 1 Find the integrating factor We form H(m)=fp(m)d:c=fadx=ax (remember, we don’t allow a constant of integration here). Then we form the integrating factor eH(m) : 6cm:- , Step 2 Multiply by the integrating factor We multiply (1) by the integrating factor e”, giving e‘m'y' (m) + eaxay (m) = ewb. We know from the theory above that we canrewrite this as d __ (2112 : (1be d3 (e W» e (If you want a proof in this special case, observe that 0+: («2% (a) z (era) y (m) + er'y (x) ). Step 3 Integrate We integrate 5; (any (a) = emb with respect to 3:, giving emy (at) = f (emb) d3: b = «em15 + C, a with C a constant of integration. Then y (2:) ea” (26” + C) 2 + Oe—aa: Cl. IE 87 is the general solution of our differential equation. Example 2 _ Solve the differential equation my’ — 2y = 35::4 subject to y (—1) 2 2. Solution There are two new features in this example. First we are given the value of y at one point, which should help us to solve for the constant of integration that we get in our solution. This type of differential equation is called an initial value problem because we are given a "starting value" for y. Another feature is that the given equation does not have the form 9’ +p(\$)y = 91(3) - So what do we do? We divide by :c : 2 I 1 3 _.._ _ 1 3; \$9 3517 () Then this has the form W y’+p(w)y = q(w) with 2 . 39(3) 2 w; and (1(3) : 33:3. (Don’t forget the minus Sign in p, it would be fatal!). Now we can proceed as usual: Step 1 Find the Integrating Factor Form How) = [pews / (-2)“ ll 2 —2lnm = 11139—2. _ Then the integrating factor is eH(:c) 2 Elm 3—2 2 55—2. Step 2 Multiply by the integrating factor We multiply (1) by m—2 : air—23f — 233—3y : 3x. 88 From our theory above, we know that we can rewrite this in the form d —2 E's-c (m 3103)) = 33:. (If you want a proof in this special case, observe that % (22-22; m) = —2a=—3y (m) + we (as) )- Step 3 Integrate We integrate % (m-zy (sun : 3m, giving \$4.1m) = [33: dm + 0 = 39:2 + 0. Then 3; (cc) = 2:2 (\$32 + 0) = 2934 + 0222. Step 4 Solve for C We are given 3; (—1) = 2, so So y<x)=—m +—:c is our solution. 89 9.2 Separable Equations A separable equation is an equation of the form 10(39) + r1(y)y’ = 0- (1) Here p (m) is a function of 3:, and q (3;) is a function of y. The equation is called separable as m and 3; appear separately. We can write this as or in ”differential form" as p(\$)da:+q(y)dy :0- Method of Solution We just integrate with respect to a: : ./(p(w)+q(y)g%)dx=/0ds=c, fp(w)d\$+/q(y)dy=0, where C is a constant of integration. We see that we treat 1; as if it is just a variable, at least when integrating. 01' Example 1 Solve as + yy’ : 0. Solution This has the form P(\$)+9(y)y’=0, where 19(93) = I and 9(9) =11, so is separable. So we integrate: [swamp/capo =>x2+y2=20 This is about as far as we can take it. We could write y2 = 20 — \$2 and then take square roots if we knew y is positive or negative. Example 2 Show that sin (.7:— 1 ) an \$23; cos y y’= is separable and solve it. Solution We want it in the form p(\$)+t1(y)y’=0- So get all the terms in y onto one side: multiplying by y cosy gives sin (cc—1) (ycosy)y’ : — 9:2 1 or _ ( 1) sm 3‘ T+(ycosy)y’=0- This is of the form 10(3) +q(y)y’ = 0, with sin (33—1) 592 and q (3;) = 900511- 29(3) = Next let us integrate: [(p(x)+q(y)%) dm=f0d\$=0 - —1 :[E—(Ede+/ycosyﬂdm=0. (*) cc das We must evaluate the integrals on the left—hand side. A substitution u : gives fimiﬁdm = /(sinu) (—1)d'u l cosu = cos —. a: 91 HIv—I Also integrating by parts, d'y y cos y E da: fycosy dy : ysiny—f(1)sinydy y siny + cos y. Then (*) becomes 1 . cos; +ysmy+ cosy = C. This is as far as we can go. Example 3 Show that y’ : may"m is separable, and solve it. Solution We have to separate the y’s and 35’s. Rewrite as y’ I (me—I) ey :> e—yy’ = 1138—”c :> —\$e’z + e‘yy’ = 0. Now integrate with respect to a: : /(—xe_m + 6—yy') d2: = dezr: :_ 0 0531 u. -I —y_.. = =>f( me )dm+/e dmda: 0 => / (_\$e—z) dsc—E—fe—y dy = 0. Here integrating by parts gives II a m I H I /—'N H V (0 I 8 5:. a II 5-} m I a + m I ”a While (*) Substituting into (*) gives ace”:E + e“: — e‘y 2: 0' => e‘y = 336—” + 8—” — 0 => —'y' = 1n (ace—x + 6'“ — C) z>y : ~1n (we—z +8“m —C). 93 ...
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