sm212 quiz 2 (1997)

QUIZ 2

SM212

PROF WD JOYNER

1) Solve (D-1)³y=0
Roots of the characteristic polynomial are : r=1,1,1

ANSWER: y=c₁e^x+c₂xe^x+ c₃x²e^x.

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=c₁e^x+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 = e^2x , y(0)=0 , y^/(0) = 0

r²-1=0

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

r=1,-1

c₁e^x + c₂e^-x = y_h

g(x)= e^2xg^/ (x)= 2e^2x

y_p =A e^2x y^/ =2A e^2x y^// =4Ae^2x

4A e^2x -A e^2x = e^2x 3A e^2x = e^2x A=(1/3)

y_p = (1/3) e^2x

y_p + y_h = (1/3) e^2x + c₁e^x + 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) e^2x + (-c₂ - (1/3)) e^x + c₂e^-x

y = (1/3) e^2x -c₂ e^x - (1/3) e^x + c₂e^-x

y^/=(2/3) e^2x -c₂ e^x - (1/3) e^x - 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)e^2x-(1/2)e^x+(1/6)e^-x

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