Recent content by mathwizarddud

anyone?

I thought that I've already used them in first determining the eigenvalue.

Consider five points A, B, C, D and E such that ABCD is a parallelogram and BCED is a cyclic quadrilateral. Let l be a line passing through A. Suppose that l intersects the interior of the segment DC at F and intersects line BC at G. Suppose also that EF = EG = EC. Prove that l is the...

Here's what I had: after solving the ODE, we have the general solution \phi = C_1 \sin(\sqrt{\lambda}x) + C_2 \cos(\sqrt{\lambda}x) applying the conditions we have the system C_2 = C_1 \sin(\sqrt{\lambda}2\pi) + C_2 \cos(\sqrt{\lambda}2\pi) C_1 \sqrt{\lambda} = C_1 \sqrt{\lambda}...

Yes, double integration is exactly what I had in mind. Try the first one; it's a bit harder. The answer is: (ln(3)+ \pi/3^{1.5})*\pi/3^{1.5}.

Solve (1-i)^n = -512 - 512i for n. :wink:

Why do (almost) everyone assume that my posts are homeworks questions? Is it invalid to post some (sensible) problems that one made up him/herself to learn the ways others solve it? LOL :-p

This is not homework of course. I know the answers already (I made one of them up), and I just want to see what would people say if they don't exactly know how to solve it (by hand) + if there's a more elegant way to solve these two so that I could learn something, too! LOL :smile:

Solve the eigenvalue problem \frac{d^2 \phi}{dx^2} = -\lambda \phi subject to \phi(0) = \phi(2\pi) and \frac{d \phi}{dx} (0) = \frac{d \phi}{dx} (2 \pi). I had the solution already, but am looking for a much simpler way, if any. EDIT: Sorry that I accidentally posted...

Evaluate \int_{0}^{\infty} x\ (e^{3x}-1)^{-1/3}\ dx and \int_0^1\frac{x-1}{\ln\, x} \; dx

Prove that a matrix A is invertible if and only if its reduced row echelon row is the identity matrix.

Find the general solution for \frac{dy}{dx} = xe^{-y} + \frac{2}{x}

Prove that if a, b, r, and s are positive reals and r + s = 1, then ar bs ≤ ra + sb.

Consider \triangle ABC with vertices A(4,8),B( - 1,2), and C(0, - 3). Find the point D such that \triangle ABD,\triangle ACD and \triangle BCD all have the same area.

Hello! I don't see how this is true: "Define w = exp(x³/3) y. Then dw/dx = xp(x³/3) x." dw/dx should be exp(x³/3) y' + x²exp(x³/3) y. Note that y = y(x).