Recent content by gbacsf

Thanks, I finished off #1. For the second question, what is meant by standard normal? Does that mean to use (0,1,0,0), (0,0,1,0) and (0,0,0,1) with the original vector in gram-schmidt to find the orthogonal basis?

I need some help in understanding what I need to do to solve these poblems, I can't get them started. 1. Find an orthogonal set of vectos that spans the same subspace as a,b,c. a=(1,1,-1) b=(-2,-3,1) c=(-1,-2,0) 2. Use the Gram-Schmidt process to find and orthogonal basis that...

Ah thanks, so e1= (1/sqrt(2), 0, 1/sqrt(2)) e2 = (2/3, 1/3, -2/3)

So set (y=1, z=0) and (y=0, z=1) Get two vectors: (4,1,0) and (1,0,1) Normalize: (4/sqrt(17), 1/sqrt(17), 0) and (1/sqrt(2), 0, 1/sqrt(2))

I need to find the Orthonormal Basis of this plane: x - 4y -z = 0 I know the result will be the span of two vectors but I'm not sure where to start. Any hints? Thanks, Gab

Is this the proper way to write the Heavenside function for the following conditions: g(t) = t if t <= 8pi 8pi if t > 8pi Heavenside: g(t) = t + (8pi - t)*U(t-8pi)

Well from that I can say that y=sin(x) is a solution. Then I get: z' +(-2*tan(x))*z = 1/cos(x) So then I solve this linear equation: (sin(x) + C)/(cos(x))^2

I'm not sure if I can solve it that way. I did some digging and found its in the form of a Riccati eq. http://en.wikipedia.org/wiki/Riccati_equation But I'm not sure how to apply it to this problem, any help? Thanks

What method would be used to solve this DE, it look like a Bernoulli but isn't. I'm lost. y'cosx = 1-y^2 Thanks, Gab

Thanks! Right, I get 1/[(-2+5z)/(2+z)-z] dz = (1/x) dx so 3ln(z-1) -4ln(z-2) = lnx +c and 3ln(y/x -1) -4ln(y/x-2)-lnx = c ?

What approach should be used to solve the following DE: dy/dx= (-2x+5y)/(2x+y) Find an integrating factor and solve it as an excact equation? Thanks.

So... Yg= u1*e^(-3.2x)+u2*x*e^(-3.2x)+0.5*(X^2)*e^(-3.2x) ?

uh? Variation of parameters to get something like Yg=u1y1+u2y2+c1(x)y1+c2(x)y2 where Yh= u1y1+u2y2 and Yp=c1(x)y1+c2(x)y2 ?

Ops, yes I do.

I'm have a lot of trouble trying to find the general solution to the following D.E. y'' + 6.4y' + 10.24y = e^(-3.2x) I get the homogeneous solution as C1*e^(-3.2x)+C2*x*e^(-3.2x) and the particular solution as 0 So a general solution of Y=C1*e^(-3.2x)+C2*x*e^(-3.2x) I know my solution is...