I need all parts to question 2 please Thank you Consider the

I need all parts to question #2 please. Thank you!

Consider the 3^rd order homogeneous linear differential equation for y(x) y\"\' (x) = 0 and let W be the solution space. By taking successive anti-derivative to solve this differential equation. Interpret your results using vector space concepts to show that the functions y_0 = 1, y_1 = x, y_2 = x^2 are a basis for W. What is the dimension of W? Show that the functions z_0(x) = 1, z_1(x) = x -1, z_2(x) = 1/3(x -1)^2 are also a basis for W. Use a linear combination of the solution basis from part (b), in order to solve the initial value problem below. y\"\' (x) = 0 y(1) = 3 y\'(1) = 4 y\"(1) = 5

Solution

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.
SO, y is a linear combination of 1,x,x^2

Since, D,B,C are arbitrary so any linear combination of 1,x,x^2
is a solution

HEnce solutin space is P2 ie 1,x,x^2 form basis for W

b)

z0=1

z1+z0=x

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

2z2=x^2+1-2x

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

2z2+2z1+z0=x^2

So,

y=Dx^2+Bx+C

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

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

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

B=-1

D+B+C=3

5/2-1+C=3

3/2+C=3

C=3/2

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

y=2.5(2z2+2z1+z0)-(z1+z0)+1.5z0

y=5z2+4z1+3z0