Recent content by scud0405

Oh, yeah, whoops. It is: v(t)= -9.8t - 4. Okay, now I make the problem: ds/dt = -9.8t - 4 Am I on the right track now? I am suppose to integrate next, correct?

Okay, so would it just be v(t)= -9.8t + 4 and then I substitute v(t) with ds/dt?

Hmm.. so to find v, I would integrate dv/dt, which would give me: v = 9.8t + C? Then, to solve for c, I do 4 = c and that gives me the equation for time which is v(t) = 9.8t + 4? Is this correct so far, or what do I need to reconsider?

Hmm... it seems it would be, but how would I set-up the formula to determine where the object is at time t? I can't do that with just dv/dt = 9.8, can I? The at time = 0, velocity is 4, so assuming no air resistance, does that mean it takes 125 seconds to reach the bottom, or does gravity alter...

Homework Statement I've been given a problem that an object falls from a distance of 800 feet with an initial velocity of 4 f/s. Gravity acts on the object, but air resistance can be disregarded. Find the location at time t and find when the object hits the ground. The Attempt at a...

Ahhh! Finally, it makes sense! Thanks!

e^\int^-^1^/^x = e^-^l^n^x = -x. So this means I would multiply the whole problem by -x which gives me what I start out with. Or am I doing something wrong?

Okay, so if I divide by X, I get y' - \frac{y}{x} = 0 So the integrating factor would be -x? That can't be right because that just gives me what I started out with.

Right. But is that the equation correctly separated?

Oh.. you're right. It's not separable though, is it? When I try and separate it I cannot get the Ys on one side and the Xs on one side. Unless \frac{dy}{y} = \frac{dx}{x} is the equation correctly separated.

Oh, okay! How did the 8c become only c in: \frac{8}{5} ln(x)+ 8c= ln(x^{\frac{8}{5}})+ c Thanks for the help.

Oh, I understand! Just multiply by e^-x, so the problem looks like this: -ye^-^x + xe^-^xy' = 0 Therefore: xe^-^xy = c Correct? I'm having trouble with the next problem also. The next problem is similar, but with a x^3 added to it. I can't think of any other way to do it other...

Homework Statement -y + xy' = 0 and y(2)=5 The Attempt at a Solution This first part trips me up. I am supposed to find the perfect derivative, which is (xy') = 0 ? Is this legal, or does the -y not allow for that? If that is correct, then I know that I integrate that, which...

So, here's the problem: \frac{dy}{dx} = \frac{8y}{5x} To start off, I separate the integrals, which gives me: \frac{dy}{8y} = \frac{dy}{5x} After that, I integrate both sides, which gives me: \frac{ln8y}{8} = \frac{ln5x}{5} + c Now, the question says that it runs through (4, 1), so that...