Recent content by Vol

Oh, ok. So, the answer is tan2x = 1. I couldn't see you have to divide both sides by cos2x to turn it into tan2x. Thanks y'all.

sin2x = 2sinxcosx and yes it's for solving for x. By answer I meant the next step.

4cos2x = 8sinxcosx 4cos2x - 8sinxcosx = 0 Now I am stuck. I don't know what identities to use. I can see it was set to 0 for a reason. But why? I know the answer is 4 - 4tan2x = 0 but how? Thanks.

h(x) = cf(x) + kg(x) is the linear combination of functions. What makes it linear?

Oh geesh! I thought it was by a mathematician. Well, I now know better.

Oh, I see. He was just thinking backwards. Thanks.

Yikes. But it is from a university site. Now I am really confused. But where did ##e^{\frac{-y^2}{2}}## come from? How did he come up with that guess?

That is exactly my point. Looking at the given equation and seeing that ##\frac {2E} {hw}## is negligible compared to ##y^2## you are supposed to guess the general solution... that ##\Psi(y)## will be as ##e^{\frac {-y^2} 2}##. Totally lost. Here is where I found this problem...

Anybody? Am I missing something altogether?

OK, so I get: ##\frac {d^2\Psi(y)} {dy^2} + (\frac {2E} {hw} - y^2)\Psi(y) = 0## Then, ##\Psi(y) = u(y)e^{\frac {-y^2}{2}}## is the general solution. Totally lost. Where did ##\frac{-y^2}2## come from?

Thanks. I get it now. Then you cancel out all the mw/h with h/mw. I will be back with questions about the next step for a series solution.

Why do we have to replace ##\frac d {dx}## by ##\sqrt \frac {mw} h## ##\frac d {dy}## ? I can see we are changing from dx to dy because we are substituting in ##y\sqrt\frac h {mw}## for x? But where does ##\sqrt \frac {mw} h## come from? Is it the chain rule?

Why doesn't Latex work? Help.

d^2 \frac d^2{x} testing Latex

Is there a way I can preview how it looks before I post it? I clicked preview but it is still in code form.