MATH 202 Exam 2 with Solution

MATHEMATICS 202 ( 1st Semester, 1994-95 ) Quiz Ii 1) Solve the initial value problem y" + 2y' + 2y = -2c0s2x - 4 sin2x, yO)=1,y(O)=1. 2) Find the general solution of the equation x? y"+xy'-y = xe* 3) Find all values of m for which the problem y"tmy=0, y(0)=0, yi(l)=0 has non-trivial solutions and find those solutions . CI) y= ye,x” ' 3, 4) Let 0 be a solution of the equation Y~4* = 9 . satisfying y(0)= 1. Find cn as a function of n and then find y. @)) Let y; & y2 be solutions of the equation :” y" + p(x)y' + q(x)y = 0 and let w be the wronskian of y; & y2. Prove that w(x) = waexp([p(@)dt), (exp t = e'). January 5, 1995 AN Solution Quiz 1994-95 1) y"+2y'+ 2y=-2cos2x-4sin2x, y(0)=1, y'(0)=1. m2+2m+2=0 1 ml=-l-i, m2=-l+i y, =e" (e, cosx + €, sinx) Hint : yp is found by the method of undetermined coefficients, don't use the variation of parameters method or else you will face difficult integrations . 2.1 2 *Y +xy'—y = x°e* m(m-1)+m-1=0 => m =i, m, =-1 en yl 1 Ve = O\X™ + 0,x yp is determined by variation of parameters . N.B. not by the undetermined coefficients . ¢ nt 5 oe One nol olwoy cer y=x” > Y,=x bot . VY, = UY, +ihyz Up= ~ BLE Findu, & u,. uje Oy 52% w 3) y"+my=0, y(0)=0, y(1)=0 Suppose m =k =>m24+k=0 case I: k=0 =>m2=0, Therefore, "y=c, tox y(0) =0=>c, =0 y'(1)=0=>c,=0 => no non-trivial solution . case 2: k<0 > k=l m+ Ik| =0 ' A =—A4|k| = 4]k|i? 2 m,m,=+ a = t/q i y= cel + a y= —k|c,e + |k|c,e"'* ypO)=0 => oc, +e, =0 y()=0 => -|klee™! +|kle,el! =|kle"! +|kle™ = 2coshe!"! #0 >c,=¢c,=0 => no non-trivial solution . case 3: k>0 => k =|k| = ~4|k] = 4d? 2,)\K} i mm, = a = +A i yHe, cos,/|A| |x+c, sin lx y0)=0 > c= y'()=0 » e filcos fi = c, must be ¥ 0 => vlK\ e0s IK] = 0 ik] =n = 1/2, 3/2, 5/2. AT >y=c, sin(—-)* is an eigen function 4) 5) yi-4x3y =0 5 ne, x"! ~45 co x" =0 Dine, 1 n=l n=0 Dk + Wey." - 43 cx" =0 k=0 ks3 y(0) =1 => c,=) c k=3 =e, =¢,=0 y" + p@xdy' + q@)y = 0 W= 19 - Yo¥1" W'= pg! * Yg"¥y~ Yy'Ya' Voy" = Ya" > Yay" w'= y,(—p(x)y'.-4(*)¥2) + ¥2 (Pp) y's +40)1) w= —p(x)WY'2-Y2Y's) w's= — p(x)w w'+ p(x)w=0 Inw+c=— | p(x)dx Inw+c= - | p(tdx x ~f par wx=e'e 7 forx=a => e=wa 6/