Consider the 3rd order homogeneous linear differential equat

Consider the 3^rd order homogeneous linear differential equation for y(x) y t prime (x) = 0 and let W be the solution space. Use successive anti differentiation to solve this differential equation. What three functions y_0(x), y_1(x), y_2(x)(arranged from lowest degree to highest) span the solution space? Why are they linearly independent ? Do they form a basis? Show that the functions z_0 = 1, z_1 = x-1, z_2 = 1/2 (x-1)^2 are also a basis for W. Use linear combination of the solution basis from part (b) to solve the initial value problem below. Y t prime (x) = 0 y(1) = 3 y\'(1) = 4 y\"(1) = 5

Solution

a)

Integrating gives

y\'\'=2A (Choose constant to be 2A instead of A)

y\'=2Ax+B

y=Ax^2+Bx+C

y0=1,y1=x,y2=x^2

b)

zo=y0

z1=x-1=y1-y0=y1-z0

z1+z0=y1

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

2z2=y2-2y1+z0

2z2-z0=y2-2(z1+z0)

2z2+2z1+z0=y2

Hence, y0,y1,y2 lie in span{z0,z1,z2}

HEnce, z0,z1,z2 form basis for W

c)

y=Ax^2+Bx+C

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

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

y\'\'(1)=2A=5

Hence, A=5/2, B=-1

C=3/2

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

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

y=5z2+4z1+3z0