MATH 202 Exam 1 with Solution

MATHEMATICS 202 ( 2nd Semester , 1992-93 ) Quiz Time : 60 mins. March 27 , 1993 1) Find the general solution of : (2x° - y+x? siny)dx + (x* cosy + x)dy = 0 (10 pts.) 2) Consider the differential equation : eAQ wx?) 2ww'+x = —.—_—-___..... (1 * x? +] (1) a) Use the substitution y= e” Bernoulli type equation (2) . to transform equation (1) to b) Solve (2) and deduce the solution of (1). c) Find the part solution that satisfies w(0) = 0 (10 pts.) 3) Find the general solution of the differential equation : y" - 3y' + 2y = cos(e*) (10 pts.) 4) Given the initial value problem : y't2y2-3y+1=0; - y(0) = 3/2 vw () Provide the solution of (1) , that must not contain any arbitrary constant . Hint : Find the constant k, so that the substitution y = kz'/z will produce a ’ second order differential equation any Math 202 student may handle . (10 pts.) 50 5) Find the.general solution of the following integro-differential equation : yit4y +3 [y(adt = sin?x + e* 0 Hint: Take the derivative of both sides . 1) (2x? - y+x? siny)dx + (x* cosy + x)dy =0 2) Solution Quizl 1992-93 ——=-l+x'*cosy Oy ON 33? cosy +1 Ox Then pt = 1/x2 is an integrating factor . uN OuM _ gy 1 Oy Ox x Exact equation : f(x,y) = f(2x- y+ siny)dx = x? atys xsiny+ p(y) x x 6 f(x, Y) 0 sy (1/x) + xcosy + g'(y) = xcosy + 1/x => g(y)=0 = gly)=e => (1/x) + xcosy =c, e3sQw sx?) 2ww'+x =—>——__.... (1 . x +] (1) y=e” => y'=2ww'e” rm -3/2 (-3/4)x2 Yyyay * which is Bernoulli y x +1 ' -(3/4)x? y v2, & yay x= x41 3 3 3 e 4 V2 ' -V2.1 , . z=y => ge-y “y'> Z2+—uxH= linear. 2 2 x? +1 substituting and solving we get : f2xdx Az=e’? integrating factor 3a => ze! = Ja dx x? +1 AL: 3) 4) 2 3 z ex =5 Arctgx +¢ ze 3Arctage +2 . Ie eos ew _ 3Arctagy c w(0) = 0 2@GIe ele? ez _ 3Arctagx Qe y" - 3y' + 2y = cos(e*) = f(x) y"-3y'+ 2y =0 is an auxilliary equation : m2-3m+2=0 m=1 & m=2 y=cje*+c,e% w=e* Wromst 2x -x v= -e F(x) | w=" f(x) w w particular solution : u'; =e cos(e*) u', =e™* cos(e™) > ue=sinfe*) & wu = Je* (e™* cos(e™))adx solving for u, by integration by parts we get : v=e™ du=e™ cose “dx u, =e * sine™* — cose * Vp =UY, + Uy, =e" sine™ +e°*(e™ sine — cose™* y'+2y2-3y+1=0; _ y(0) = 3/2 kz' oe ' vA Ke de® Zz 2 4 12 G fa te oye 3 4 120 Zz 2 Zz Zz vA : 2k?-k=0=>k=Ouneccaptable, k=1/2. Let k= 1/2 Lin 3 ae =0 2 2 m —3m+2=0 m=1 & m=2 ie" +2C,e7*) 1¢ z=ce%+c,e% = y=kZ ibe , 2 Cye* +c,e* Cie" +C,e -—2___. y(0) == y C,e* +C,e”* 2 Cc 3 > +C, 2s =C,+2C, =3C,+3C, > C, =-2C 2 C,+C, y’=-2y2 +3y -1 y’(0) = -2(9/4) + (3x3)/2 -1 =-1 By substitution we get another equation. 5) x yit4y + 3 [y(e)de =sin’x +e" 0 y'" + 4y'+ 3y =2 sinx cosx - ex y" + 4y'+ 3y =sin2x - ex This is a homogeneous equation y=ce* +c,e™ ¥, = Asin2x + Bcoos2x + Ce™ A,B,&C can be found by the method of undetermined coefficients