MATH 202 Exam 1 with Solution

MATHEMATICS 202 (1st Semester, 1994 - 95) Time: 60 mins. QUIZ I Novembe 24,1994 . l. Solve the following: (a) (1+ 2eX)dx - (1 + e%)(2y + cosy)dy = 0; y(0) =0. (b) (y - yx2-y? ) dx - xdy = 0 dy 2y_. 2 (c) dx x (d) o = x + y using the substitution y = ve -dx 2. Determine A so that (3x2y2 + Axy)dx + (2x3y + 2x2)dy =0 is exact. If y is a solution of the resulting equation satisfying : y(1) =2 and y(2) > 0, find y(2). 3. If y =u + x satisfies the differential equation dy , i _ dx + x = xX, find u. 4, If y =x and y = sinx are solutions of y" + ay' + by =0, determine the functions a and b. Ye. 1) a) b) Solution Quizl 1994-95 (1 + 2eX)dx - (1+eX)(2y + cosy)dy = 0 x + => 1+ 2c" dx = (2y + cosy) dy lt+e x 4x ae dx = (2y + cosy) dy e x q+ e ) dx = (2y + cosy) d 1 + @* é x +Ln(1 + eX) +C=y2 + siny y(0) =0 => 0+Ln2+C=0+0 > C=-bLn2 y2 + siny = x + Ln(1 + eX) - Ln2 (y --y¥x2-y*) dx - xdy = 0 y=ux => dy =udx+ xdu (ux - Vx2-u2x2) dx - x(udx + xdu) = 0 x(u - v1 ~ 2) dx - x(udx + xdu) = 0 > - vi - w2 dx - xdu = 0 du | & 6 arcsim = - Ln|x| + C 1-- u2 x => u = - sin(Ln|x| + C) dy + -2y = xy2 dx x => y dy + 2 y! =X dx x -1 dw -2 dy Let w = > WF >. y dx y dx : dw 2 => -— —w =x dx x dw 2 : . — --—w = -x Linear equation. dx x 2 m= fax = x? 2 dw 2 _ -] 7 a eX d, 1 — WwW => - — ax * ) x x?w = -Lnx +C >w=--U™% poze y ‘ x2 4G d) 2) y = ver 1 dy dv > —-— =v xX + Xm dx dx dv vex + eX — = xt ver dx dv => eX — = dx => dv = eX xdx v= J(e*% xdx) =-xe7X + JeXdxC =x dv X u du = dx Vv ott € -e = -xe7X - e™%* +C= Ve >y=-x-1+C = -x+C xX (3x2y2 + Axy) dx + (2x3y + 2x2) dy =0 OM _ 6x* + Ax dy ON Ox. >A=4 0 f(x,y) _ Ox f(x,y) = x3y2 + 2x2y + g(y) = 6x" y + 4x 3x’y° + xy of ay >eg(y)=0 ~gy=C = 2x3y + 2x2 + gy) = Yt => f(xy) = x3 y2 +2x2y +C= y(1) =2 > Cz =-(44+4)=-8 x3y2 + 2x2y -8=0 y(2)=8 y2 + 8y-8 A = 64 +9(64) => JA =845 8+ 8/5 _ -1+ V5 y(2) = = 16 2 dy 2 Uy dx dx ge iyylay dx x 1 => du =(x- — - I)dx x x2 w= 7 Lal x + y=x y = sinx y" tay! + by =O vse (0) y=x > y=1&y"=0 (y(2) > 0) => a=-bx y=sinx > y'=cosx & y" =-sinx - sinx + a cosx + bsinx = 0 - sinx - bx cosx + bsinx = 0 => -sinx + b(sinx - xcosx) = 0 sin X => b= ——— _ sin x - xcos xX sin x a= -x sin X - xcos X E%