MATH 202 Exam 1 with Solution

lL. (a) (b) 2. (a) (b) 3. (a) (b) (c) MATHEMATICS 202 (2nd Semester, 1993-94) March 26, 1994 QUIZ 1 1 Use an appropriate substitution and solve: (x+2y-1) dx + 3(x+2y) dy =0 Subject to: y(1)=-1 Solve the initial value problem: y(x2+y2)dx + x(3x2-S5y2)dy =0; y(2)=1 (8+ 10 = 18 pts) © Find r so that x! is an integrating factor of the equation: (3xy2+4y)dx + (4x2y + 8x)dy =0 then find the general solution. Solve: y=y2-y-2; y(0)=1 (9+9= 18 pts.) Given the differential equation: y" + Ay! + By =0 wee (1) Find constants A and B so that y1 = exsinx is a solution of (1) Find a second linearly independent solution y2 of (1) Using the result of parts (a) and (b) above, determine a particular solution of : by inspection. Then detemine the general solution of (2). ‘(3+6+3=14 pis.) 3H b) Solution Quizl 1993-94 1) (xt2y-1) dx + 3(xt2y) dy = 0 y(1) =- Letu=x+2y => du 42 => dx dx u-1 + 3/2(u)(u'-1) = 0 3/2(uu’) - 1/2(u) -1 = 3uu'-u-2=0 3udu - (ut2)dx = 0 3M y= de u+2 => JS au= xte integrating we get : 3u —6lnlu+2|\=x+e 3(x + 2y) — 6ln|x+2y+2)=x+e yd)=-l => =-4 => 3(x +2y) —6lnlx+2y+2)=x-4 y(x2+y2)dx + x(3x2-5y2)dy =0; y(2)=1 This is a homogeneous equation of order 3 . Lety=ux => dy=xdu+udx ux(x? + u?x?)dx + x(3x? —5u?x?)(xdu + udx) =0 xeu(ut Ide +x*(3—5u")du + x°(3—5u?)udx =0 x (-5u° + u> + 4u)dx + x‘ (3 —5u)du 5u’ -3 Su = = In|x|+c 5 +0? +4u by partial differentiation we get : 2 - f ou 3 ux {( 4 _ lAue 4/3, 4, u(—S5u tus 3u Su? +u+4 integrating we get : Injx| += In — in °Y 4244) linl 3 |x} 12 | x x 3 |x it is easy to find c by replacing x & y by (2 & 1). 2) a) b) u=x!l (3xy2+4y)dx + (4x2y + 8x)dy =0 ux M=3x"!y? +4x'y r+i Bx N= 4x?*y 48x @NM = 6x"! y 44x" dy aux = 4(2+1r)x"!y +8(r+1)x x 6x" y+ 4x" =4(2 +r) x" y + B(r + I)x 6=8+4r>r=— 2 verification : 4=8(-1/2+ 1) (3x1? y? + 4x7 y)dx + (42°? p + 8x"? )dy =0 Exact — Equation [ Max =2y?x*? + 8yx'? + gy 0 (Max) _N oy => 4yx®? + 8x!" + o(y) = 4x7? y + Bx!” gy) =0=> gy) =4 Qy?x*? + 8 yxl? + c, =¢, y=ye-y-2 y(0)=1 y'ty-y2+2=0 Ricard, Von bose Guctlik. Lety=y,+u y, =2 y=2+u you oy =>u'-u + 3u=0 Bernoulli ¥3.y Let z= I/u u u => "= -u'/u2 z-3z+1=0 linear equation N= e-3x l1 ec 1 1 Z=-—+ ax = - = 3 e uoy-2 ; substituting values of x & y we get: c=-4/3 3% 3) b) y" + Ay' + By=0 yl =eXsinx => — y'=eXsinx + cosx(ex) y" = 2eXcosx 2e%cosx + A(eXsinx + cosx(ex)) + Bexsinx = 0 A=-2 A=-B => B=2 y"-2y'+2y=0 : ~|-2de V2 =); ‘- 7 dx = —e* sinxcot gx (wy) y"-2y'+2y=1 y = 1/2 is a particular solution *sinx + ¢,e* Si t -¢teo ce SiInx + Cc,e SINX COL gx = (General solution)