Consider the 3rd order homogeneous linear dierential equation for y(x) y (x) = 0

Question

Consider the 3rd order homogeneous linear dierential equation for y(x)

y (x) = 0

and let W be the solution space.
2. (a) Use successive antidierentiation to nd the general solution of this dierentialequation. Interpret your results using vector space concepts to show that the func-tions y0 = 1, y1 = x, y2 = x2 are a basis for W. What is the dimension of W?
(b) Show that the functions z0 = 1, z1 = x 1, z2(x) = 1

2(x 1)2 are also a basis for W.
(c) Use linear combination of the solution basis from part (b) in order to solve theinitial value problem below.
y (x) = 0y(1) = 3y (1) = 4y (1) = 5

Explanation / Answer

a)

y'''=0

Integrating gives

y''=A

y'=Ax+B

y=Ax^2/2+Bx+C

Choose, A/2=D

y=Dx^2+Bx+C

D,B,C are arbitrary constants

HEnce, any linear combination fo 1,x,x^2 is a solution

Hence, 1,x,x^2 are a basis for W

dimension of W=3 as basis has 3 vectors

b)

z0=1

z1+z0=x

z2=(x-1)^2/2=x^2/2-x+1/2=x^2/2-(z1+z0)+z0/2

2z2+z1+2z0-z0=x^2

2z2+z1+z0=x^2

Hence, z0,z1,z2 contains 1,x,x^2 hence, span W

c)

y=Dx^2+Bx+C

y(1)=3=D+B+C

y'(1)=2D+B=4

y''(1)=2D=5 , D=5/2

B=-1

D+B+C=3

5/2-1+C=3

C=3/2

y=5x^2/2-x+3/2

y=5(2z2+z1+z0)/2-(z1+z0)+3z0/2

y=5z2+3z1/2+3z0

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