Recent content by physicsfan24

Its late...and I've been bashing ny head against my desk thinking about this problem

Sigh...ok so, |e^kt-e^-kt|<=2Me^ct |e^kt-1/e^kt|<=2Me^ct |(e^3kt-e^kt)/(e^2kt)|<=2Me^ct |e^3kt-e^kt|<=2Me^[t(c+2k)] ... ln|e^3kt-e^kt|<=ln|2Me^[t(c+2k)]| ln|e^2kt - 1|+ tk <= (c+2k)*t+ln|2M| ln|[e^2kt - 1]/(2M)]| <= (c+2k)t+kt ln|[e^2kt - 1]/(2M)]| <= (c+3k)t OK...? uhh... now...

Homework Statement Prove that sinh(kt) is of exponential order |k|. Find M>0, c>=0 and T>0 to show that |f(t)|= Me^(ct), t>T Homework Equations |f(t)|= Me^(ct), t>T The Attempt at a Solution I'm looking at the graph of sinhkt (i graphed a few values of k) and indeed it is of an...

Umm, I'm in U of Virgina and this is an online HW question... Yes I am taking ODE. You're in my class?

Homework Statement This is the original question: \frac{d^{2}y}{dx^{2}}-\frac{6x}{3x^{2}+1}\frac{dy}{dx}+\lambda(3x^{2}+1)^{2}y=0 (Hint: Let t=x^{3}+x) y(0)=0 y(\pi)=02. The attempt at a solution This might be all wrong, but this is all I can think of \frac{dt}{dx}=3x^{2}+1 so...

Ok so it would be 2 * \int^{1}_{0} \frac{1}{(1+x^4)} * \frac{1}{(2u^1/2)}du Now, is my logic correct when I do? u = x^2, so x= u^1/2 dx= 1/2 u^(-1/2)

Question: \int^{1}_{-1} \frac{dx}{(1+x^4)} I attempt: u = x^2, so x= u^1/2 dx= 1/2 u^(-1/2) Which gives me \int^{1}_{1} \frac{1}{(1+x^4)} * \frac{1}{(2u^1/2)}du, which is 0. Thats not the answer as seen by any graphing utility. Where is this error? I do not know integration by parts. I just...