Recent content by EtherealMonkey

Sorry, I don't know what photobucket did to my png. Here is the same image in PDF format (only marginally better - IMHO, but it is readable w/zoom :smile: )

Okay, here is a screen-shot of Maple with the problem loaded. The issue that I am inquiring about occurs at Step #9 on the image below. As I stated in the first post, I am re-working this problem and cannot determine why I was able to find f(a)=13524.29951 when I originally worked the problem...

This is a problem that I am working on in Maple 12. I have assigned the Differential Equation and the Initial Values for the two parts. Then, I assigned the result to a variable before splitting off the two functions that return from dsolve. Plotting the first function and the initial...

So, this is where I am stuck: ln\left(y\right)+y^{2} = \sin{x}+c_{0} I am confrused... :blushing:

Homework Statement Homework Equations I think that I need to find the moment of the force F about the line BC first. Then, using that moment, find the "projection" of M_{BC} onto the X axis to find the answer. M_{BC}\sin \left(45^{o}\right) However, I am not getting anything...

Yes, thank you very much! Sorry for the late response, I just got on campus. Now, got to go take a test, wish me much success! Thanks again Mark!

This is what I have done so far... P\left(t\right)\frac{d}{dt}=P\left(t\right)-P\left(t\right)^{2} P\left(t\right)\frac{d}{dt}=P\left(t\right)\left(1-P\left(t\right)\right) \frac{1}{P\left(t\right)}dP=\left(1-P\left(t\right)\right)dt \int \frac{1}{P\left(t\right)}dP= \int dt - \int...

Please? I know this may seem so obvious. And, in fact a great deal of my trouble in D.E. so far can be summarized by my "not seeing the forest for the trees". So, any hint would probably get me going...

Homework Statement \frac{dP}{dt}=P-P^{2} It seems that Partial Fractions should be used to solve this D.E., but I cannot find an example to go by. I even tried to rewrite the equation as: \frac{d}{dx}Y\left(x\right)=Y\left(x\right)-Y\left(x\right)^{2} But, that isn't helping me...

Yeah, my CalIII is killing me... Thanks for the response. I hope the next time I have a question, it will be a good one :redface:

Okay, never-mind... Problem: \frac{dy}{dx}+2xy^{2}=0 Solution: \frac{dy}{dx}=-\left(2xy^{2}\right) \left(\frac{1}{y^{2}}\right)\frac{dy}{dx}=-2x \int\frac{1}{y^{2}} dy=-2\int x dx y=\frac{1}{x^{2}+C_{1}}

Homework Statement \frac{dy}{dx}+2xy^{2}=0 I am stuck on this. I realize that this is a non-linear exact equation, but I just cannot wrap my mind around any type of method to attack this one. TIA for any help

I started to try to solve this myself and drew the same on paper. Thanks for posting the link with the method for solution. I couldn't figure it out either.

Jay, thank you sir!

The problem statement: My relevant equation: \phi will be the angle between the X axis and F_{CO} \theta = \phi + \arcsin\left(\frac{3}{5}\right) My attempt at a solution: \Sigma F_{x} = 0: F_{CO}\cos\phi - F_{BO}\frac{4}{5} = 0 F_{CO} = \frac{F_{BO}\frac{4}{5}}{\cos\phi} \Sigma F_{y} =...