QUIZ 2

SM212

PROF WD JOYNER

1) Solve (D-1)3y=0

Roots of the characteristic polynomial are : r=1,1,1

ANSWER: y=c1ex+c2xex+ c3x2ex.




2) Solve y// + 4y = 0, y(0)= 0, y/(0)= 1

y2 + 4 =0 y=c2sin2x

r= 2i, -2i y/ = +2c2cos2x

a=0 b=2 y/(0)=1= 2c2cos2(0)

1=2c2

c1cos2x + c2sin2x = y c2=1/2

y(0)=0=c1cos2(0) + c2sin2(0)

0=c1



ANSWER: y=(1/2)sin(2x).




3) Solve (D3 + 2D2 - D - 2)y=0

(D3 + 2D2 - D - 2) / (D-1) = D2 + 3D + 2

(D-1)(D2 + 3D + 2)=0

(D-1)(D+2)(D+1)=0

D= 1,-2,-1

ANSWER: y=c1ex+c2e-2x+ c3e-x.







4) Solve (D2 + 1)2y=0

(D2 + 1)(D2 + 1)=0

D=i,i,-i,-i a=0 b=1

ANSWER: y=c1cos(x)+c2sin(x)+ c3xcos(x)+c4xsin(x).




5) Solve y// - y = e2x , y(0)=0 , y/ (0) = 0

r2-1=0

(r-1)(r+1)=0

r=1,-1

c1ex + c2e-x = yh

g(x)= e2x g/ (x)= 2e2x

yp =A e2x y/ =2A e2x y// =4Ae2x

4A e2x -A e2x = e2x 3A e2x = e2x A=(1/3)

yp = (1/3) e2x

yp + yh = (1/3) e2x + c1ex + c2e-x = y

y(0) = 0 = (1/3) e2(0) + c1e(0) + c2e-(0)

0= (1/3) + c1 + c2

c1 = -c2 - (1/3)

y = (1/3) e2x + (-c2 - (1/3)) ex + c2e-x

y = (1/3) e2x -c2 ex - (1/3) ex + c2e-x

y/=(2/3) e2x -c2 ex - (1/3) ex - c2e-x

y/(0)= 0 = (2/3) e0 -c2 e0 - (1/3) e0 - c2e-0

0 = (2/3) -c2 - (1/3) - c2

2 c2 = (2/3) -(1/3) = (1/3)

c2=(1/6)

c1 = (-1/6) -(1/3) = (-1/2)


ANSWER:
y=(1/3)e2x-(1/2)ex+(1/6)e-x

  

Written by MIDN 3/c Bernice Javier on 28 April 1997.