Recent content by byrnesj1

ahh that works beautifully. thanks ehild.

Homework Statement Find a solution (Z2) of: z'' + 2z - 6(tanh(t))2z = 0 that is linearly independent of Z1 = sech2Homework Equations The Attempt at a Solution reduction of order gives you v''(t)(Z1(t))+v'(t)(2 * Z1'(t)) + v(t)(Z1''(t)+p(t)Z1'(t)) = 0 however the third term on the LHS can be...

Thank you! Also I was attempting to simplify in terms of sinh(t) and cosh(t).. I don't know why.

I would assume that if s => ta /b, g(s) => g(ta/b) if a,b are arbitrary / real constants. I could be wrong though.

yeah.. I keep getting stuck. currently at: \frac{[-6sinh(2t)cosh^2(t)+4cosh(2t)sinh(2t)+2sinh(2t)-2sinh^3(2t)]}{(cosh(2t)+1)^3}

problem 1. you need to specify that all card are different and singular. i.e (13c1 * 4c1) (12c1 * 4c1) ... denominator is good and so is (1 - .. ) problem 2. choose a suit (4c1), now you need 5 specific cards. (13c1)*(12c1)*(11c1)*(10c1)*(9c1).. denominator is still (52c5). should get a...

Homework Statement Show that w(t) = tanh(t) solves the nonlinear problem: w''(t)+2w(t)-2w3(t) = 0 t ε ℝ Homework Equations \frac{d^2tanh(t)}{dt^2} = -2tanh(t)sech2(t) = \frac{-8sinh(2t)cosh^2(t)}{(cosh(2t)+1)^3} tanh(t) = \frac{sinh(2t)}{cosh(2t)+1} tanh(t)3 =...