Recent content by Locoism

Awesome, r from 0 to 1, and z from r^2 to √(2-r^2). Thank you so much, I see it perfectly now.

How so? I was thinking of splitting it into 2 parts and having it go from 0 to the function in terms of z, but then I would have to integrate r first, and I run into the same problem... ie: z from r^2 to 1, then from 1 to √(2-r^2), and r from 0 to √z and then from 0 to √(2-z^2)... Is...

Homework Statement Find the volume of the solid that lies between z=x2+y2 and x2+y2+z2=2 Homework Equations z=r2 z=√(2-r2) The Attempt at a Solution So changing this into cylindrical coordinates, I get z goes from r2 to √(2-r2) r goes from 0 to √2 theta goes from 0...

Excellent question. My teacher is a lunatic.

Yea, I've looked, but all I can see is if I were to do some manipulation to make it a Bessel function, which we haven't covered either. Apart from that, I was thinking it may be multiplied by a factor e^{-16} or e^{-4} so that what we have is really Y(s-a), but it's that s2 that's tripping...

I tried that, but what's the inverse transform of \frac{8}{\sqrt{s^2+16}} ??

Homework Statement Solve: \int_0^t y(τ)y(t-τ)dτ = 16sin(4t) The Attempt at a Solution My approach was to look at this as the convolution product y(t)*y(t), who's laplace transform should be Y(s)Y(s) = Y(s)2. (Note: Maybe Fourier series are better but we haven't covered that...

Homework Statement tx'' + (4t-2)x' + (13t-4)x = 0 Use laplace transform to solve. The Attempt at a Solution I've split up the X'(s) and X(s) and integrated to get X(s) = \frac{c}{(s^2 + 4s + 13)^2} = \frac{c}{9} \left[ \frac{3}{(s+2)^2 + 9} \right]^2 From this I'm guessing...

Ok well that would be much easier to solve. I guess I'll assume there's a typo in the question because I was asking myself the same thing. Glad to know I'm not insane after all.

Ok, I'm still not sure how that changes anything... (s+6)Y(s) - 3 = 3L(\frac{dx}{dt}x) - sX(s) +2 = 2sY(s) -6 I could solve for that middle transform, but replacing it into another equation will just give me 0=0... What now?

Homework Statement Use Laplace transform to the system: \frac{dy}{dt} + 6y = \frac{dx}{dt}3x - \frac{dx}{dt} = 2\frac{dy}{dt} x(0) = 2 ; y(0) = 3 The Attempt at a Solution I've tried everything on this one. I first solved \frac{dy}{dt} + 6y = 2\frac{dy}{dt} and I got y = 3e^{6t} ...

Lol. But what if we assume the rocket is very big ball thrown by a really strong dude? O.o As for the actual math... What? How do I do that? It seems that whenever I integrate of differentiate they will always remain since they multiply y? The only way I can see they are related is that...

Homework Statement Suppose a rocket is launched from the surface of the Earth with initial velocity v_0 = \sqrt(2gR) , the escape velocity. a) Find an expression for the velocity in terms of the distance x from the surface of the Earth (ignore air resistance) b) Find the time...

Yes that's right, but I was under the impression the integrating factor would be a function of x, seeing as when we assume ∂μ/∂y = 0, this would be false if μ = 2y, and consequently fail to make the equation exact. So can the integrating factor be a function of y, or even both x and y?

I'm taking my first ODE course, and I'm unsure what is meant in this question when it asks "Solve the following differential equation" (I have a list of DEs to solve). Some of them are really messy and I can't figure out an implicit solution. Would an explicit solution be acceptable? Example...