Recent content by smashyash

So I'm doing some practice problems to prepare for a test on Friday and I'm just curious about this problem:: y'' + 3y' + 2y = 4e^(x) in factoring using characteristics: (r+2)(r+1) = 0 r = -2,-1 so Yc = C1*e^(-2x) + C2*e^(x) y1= e^(-2x) y2= e^(-x) (skipping some algebra)..I...

Actually, I now figured out that if you divide x throughout the equation, you can get a nice bunch of (y/x) functions to substitute v with! My answer doesn't quite seem right though.. (1+v)y' + v(3+v) = 0 substitute v = (y/x) ; y = vx y' = (v+xv') (1+v)(v+xv') + v(3+v) = 0 v+xv' =...

thanks again! some of these are kind of tricky. I'm actually stuck on another now:: x(x+y)y' + y(3x+y) = 0 do you advise multiplying through these or not? I'm not very good at recognizing how to get started on these problems yet..

Thanks so much! Think I got it :)

I have the following equation:: xy' = y + 2*sqrt(xy) I know I should either use the F(y/x) substitute or Bernoulli's method of substitution but I'm not sure how to manipulate the equation to determine which it is. If someone had some helpful tips on how to start, please let me know...

Homework Statement I'm not sure if this can be answered without the picture but I can't get it posted. The problem gives theta(incident) and theta(r). It's a basic refraction through a transparent plastic piece. It askes for the index of regraction of this material. Homework Equations...

Thanks you! That's a great visual aid! This is still the trouble I have... visualizing exactly where this sphere is or what it's boundaries are.. I would think that both theta and phi are from 0 to pi/2 and I'm really unsure about p..

Oh, yes. I'm mistaken... Unfortunately, I still don't know how to finish setting up with new integral...

Homework Statement evaluate the following triple integral in spherical coordinates:: INT(=B) = (x^2+y^2+z^2)^2 dz dy dx where the limits are: z = 0 to z = sqrt(1-x^2-y^2) y = 0 to z = sqrt(1-x^2) x = 0 to x = 1 Homework Equations The only thing I know for sure is how to set...

great! thanks so much! :)

Ok, so here's the algebra I have: y^2 = A^2*exp(x/d) ln(y^2) = ln( A^2 * exp(x/d) ) ln(y^2) = ln(A^2) + x/d So if this is true, then is ln(A^2) = b and 1/d = m? y = mx + b

I'm really not sure... I don't really see how ln(y^2) relates to the equation, there's no way to manipulate the equation to get just that on one side..

Homework Statement I'm given an experimental equation: y = sqrt[A^2*exp(x/d)] (with a linear trend of ln(y^2) versus x) I am suppose to determine the values of first A, then d given a linear fit slope and a y intercept. Homework Equations y = mx + b The Attempt at a...

The only other numeric to satisfy this would be -(sqrt(3)). That's just from looking at the equation. Is this right?

Oh! I'm sorry.. this is confusing. So x and y have to be sqrt(3) to satisfy that equation, so these are my critical points??