Sol4 - MATH 138 Calculus 2 Fall 2010 Solutions#4 1L ~3=MT I am xda Qua 2 dy my «3L m[‘11(r l!I—Iy =>(II—12 => y—Ierl"950 =>

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Unformatted text preview: MATH 138 Calculus 2 Fall 2010 Solutions #4 1L. ~3=MT I am xda' Qua 2 .,, dy_ my «3L m [‘11-] ).(r+l)!I—Iy => (II—12+] => y—Ierl ["950] => y— x2+l a in Ln] = %l11(12 + l) + (7 [11 = 3:2 +1. du = 2xdx] = ln(:It2 +1)”2 + Inc" = ln(e("\/a:2 +1) :9 |y| = r" \/2~'-' + 1 => 3; = K «.72 + l. where K = :tc" is a constant. (In our derivation, K was nonzero, but we can restore the excluded case 3/ = 0 by allowing K to be zero.) ,J 7 ,, V 77 i , 7, ,4 1 (- Comparing the given equation. 31’ + 23; = 2:”. with the general form. y’ + P(;r)y = (3(1), we see that P(_1:) = 2 and the integmting factor is [(1) = efpuw’ = cI 2'" = 02’. Multiplying the differential equation by I (9:) gives e21yl+2c2xy=2831 =5 (e2ry)l=2(,8: => 82ry=f2pflrd$ => r21”: 223633.:- +0 => y: %CI+CP"21. )- u’) y’ = a: + 53; :5» y’ — 51) = 3:. 1(3) = efpmd’ = e“‘°"“ = 3'5”. Multiplying the differential equation by [(2) —5:yr _ 5e—5zy = Ire—6:: => (8-5::yy = 5:: gives e xe‘5’ => e‘s’y = fze‘szda: = ~§me" — file—5" + C [bypans] => y=—%z—5‘g+0e5’. "*\ ltd) éomoi'llfl‘; , 0M W “Wat flun/tr' fax a ‘m DE Xv- flu gent/id. fivm- North 34 {RM {sinl- 3.)]: Sinat- ‘3' 14.0qu ‘1. H-W—c, Sinx- % +£4.51)? = X‘Sln ‘96.) => %[$in1 - : 1.5M htt) =3 Smut. 3 = S ac-sinu‘wlx =“Z cos (x‘ )+& (5‘! ““L‘Hhfia’fl u: x* => -— ‘Ioos lx‘) '\- C ‘3: Sinx o dz: 2: 23:; => ydy=xdm => fydy=fzdx => éy2=éz2+a y(0)=—3 => ll NIO %(-3)2=%(0)2+C = C .sofiy2=§x2+§ =:> y2=xz+9 => y=——\/:cz+95incey(0)=-3<0. %=\/fi => dP/J'F=\fldt => fP"”dP=fc‘/’dt => 2P‘/’=-§-t3/’+C. P(l)=2 => 2¢§=§+C => C=2f—%,m2P"2=§t3/2+2\/§—§ => x/T’=§tm+x/‘~§ => P = (31?” + s/i — an. d 77““) W'+y=y’ => z—y-=y’-y => xdy=(y’-y)d1 => dy =% =~ d¢ yg-y dy __ d1: 1 l _ da: , /y(y_l)-/x W990.” => f(y_l y)dy— ? => lnly—l|-]n|y|_lnla:|+6 => y-l C y—l 6 11—1 0 I In =|n(e le) => — =e [21 => —=Km,whereK=:!:e => 1——=Ka: =? y y y U l l ' i=1—Ka: => y—1_Kz [Theexcludedcases,y=0andy=LareruledoutbytheinitialconditionyU)=-l.] Now (1)-—l => —1-—l- => 1-K-—l => K—2so - 1 y ‘ _l——K ‘ “ "‘1—23' dll 3 -- d1! 2 2 ' - - 12m" 1 DIVIdbettogctE+7fl=t’wh|ch|5hncar. =e2nl=f2_ Ml- - 2 - adv 4 2 r 4 2 15 (a 1' uuplylngbyt givest E+2ty=t => (1 y) =f =9 ty=gt +C => y=E+F_Thus’ 0—(1—l+c = c——1 J3 ‘ ‘3' *5 - 3-W-3'53 dy 2 ' ' 3T - 81 (12+1)d—:r-+3x(y—1)=0 => (r +1)y+3ry=3w => y+$2+19—12+1' ‘ 3/2 . . 2 . [(1) = (afar/(1244).}: = e(3/2)ln‘39+l| = (club-3+”) = ($2 + 1)::12’ Multiplymg by ($2 + “3/ Ewes I (12 +1)3/2y' +3:::(ar2 +1)"21 = 121$? +1)”2 => [(7.2 + 1):”231] = 32::(m2 +1)”2 => (1‘2 +1)”2 y = f3x(:r2 + 1)”2 dr == (:172 +1)”2 + C => y = 1+ C(zr2 + 1Y3”. Since ya!) = 2, we have 2=1+C(1) => 0: landhencc.y=1+(r2+1)'m- ...
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This note was uploaded on 11/25/2010 for the course MATH 138 taught by Professor Anoymous during the Spring '07 term at Waterloo.

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Sol4 - MATH 138 Calculus 2 Fall 2010 Solutions#4 1L ~3=MT I am xda Qua 2 dy my «3L m[‘11(r l!I—Iy =>(II—12 => y—Ierl"950 =>

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