# Search results

1. ### Easy stuff, i'm just retarded

Well you know that by the properties of integrals, you can treat it as two integrals I=I_1+I_2=\left(\int 1dt\right)+\left(-\int\cos{t}dt\right) Do you know these integrals? What is the derivative of -\sin{t}?
2. ### Find the equation of the line of intersection of the planes:

There are two parts two a problem like this. First you'll want to find the direction of the line of intersection, which is nothing but the cross product of the normal vectors of the planes, i.e. \mathbf{n_1}\times\mathbf{n_2}=\langle 2,-1,-1\rangle \times \langle 1,2,3\rangle. Then all you need...
3. ### Derivative Problem

No problem. Watch out though, I think you have a sign error in there. f'(x)=-e-x+2e-2x=0 So then 2e-2x=e-x and therefore by dividing you get 2=e-x+2x
4. ### Derivative Problem

Double check your derivative, you shouldn't be bringing an x down right? The derivative of ecx for a constant c is just cecx
5. ### Need help with differential eq. from book

To solve differential equations in the form y'+P(x)y=Q(x) it is useful to use an integrating factor defined by \mu=\exp{\left(\int P(x)dx\right)} We multiply both sides of the equation by this, \mu y'+\mu P(x)y=\mu Q(x) and if you look closely the left hand side is the...

7. ### Laplace transform of sin(2t)cos(2t)

Well if you have that \sin{(4t)}=2\sin{(2t)}\cos{(2t)} then \sin{(2t)}\cos{(2t)}=\frac{1}{2}\sin{(4t)} So then if you want to do it from the integral you just need to integrate \int_0^{\infty}\frac{1}{2}\sin{(4t)}e^{-st}dt where the s you treat as a constant (you're...
8. ### Laplace transform of sin(2t)cos(2t)

Consider that \sin{(4t)}=2\sin{(2t)}\cos{(2t)}
9. ### Limit problem

Have you tried anything yet? Where does the expression tend to as x goes to -1?
10. ### Using Laplace Transforms to solve IVP's

You've pretty much finished it, all you need to recognize that Y(s)=\frac{s-1}{(s-1)^2+1} through completing the square in the denominator. Now can you get that to work with f(t)=L^{-1}\left\{\frac{s-a}{(s-a)^2+b^2} \right\} = e^{at}\cos{(bt)}
11. ### Max & Min

You are right about the local max and min. For the second part--all an inflection point is, is where the second derivative (that which we use to determine concavity) is 0. So you would have f''(x)=2(x-2)+4(x+1)=0 Can you get it from there?
12. ### Differentiating a circle

Usually with cases like this where it is inconvenient to differentiate explicitly you can use implicit differentiation.
13. ### Greens theorem-help setting up correct integral

When you integrate a region, you do so on an interval x\in [a,b],y\in [c,d]. Does the integral make sense if a>b, c>d?
14. ### Greens theorem-help setting up correct integral

It looks fine except I would check your order of limits--it should be entering at x=0 and leaving at x=y-2 and it's backwards for y too--maybe that was just a mistake when you wrote it.
15. ### Help whith fourier transform

Oh ok I see what you're getting at--I haven't taken my complex analysis course yet, but that argument is very interesting.
16. ### Help whith fourier transform

Don't go back to sin and cos it's much easier to deal with the exponential terms. Sorry I didn't mention it but it looks like they are Gaussian, you want to get it in the form \int_{-\infty}^{\infty}e^{-(x-a)^2} where this is a Gaussian integral centered at a. So you're dealing with right now...
17. ### Finding Maclaurin Series Expansions of Functions

Yea it does take awhile; it looks like the first 4 non-zero terms are (it takes 21 terms to get them) 1-4x^7+10x^{14}-20x^{21} That looks like \sum a_n(-1)^nx^{7n} where the [itex]a_n[/tex] is some sequence that makes the {1,4,10,20,...}; I can't think of what would work for that at...
18. ### Finding Maclaurin Series Expansions of Functions

Did you find the first 4 terms of the MacLaurin series for it yet? The idea is to look for a pattern in those to find the general form of the series.
19. ### Sum of Geometric Series

Notice that 1-x+x^2-x^3+\ldots=\sum_{n=0}^{\infty}(-1)^nx^n=\sum_{n=0}^{\infty}(-x)^n
20. ### Inverse of This Laplace Function

\frac{1}{(s^2+9)^2}=\frac{1}{s^2+3^2}\frac{1}{s^2+3^2} Can you make that a convolution of two inverse transforms you already know?
21. ### Find a vector parametrization for: y^2+2x^2-2x=10

Well this basically looks like an off-center ellipse as you can see from the x term, so you'll want to parametrize it using the trig identity \cos^2x+\sin^2x=1 What happens if you complete the square on the x term and make it as such?... y=\sqrt{\frac{21}{2}}\sin{t}...
22. ### Tough Integral

I think there is an error in your polynomial division if what you are meaning to say is that \frac{2x^3}{x^3-1}=2+2\frac{1}{x^{-1}}=2+2x The way that I would do it is to split up the integral like this.. \int\frac{2x^3}{x^3-1}dx=\int x \frac{2x^2}{x^3-1}dx=x\ln{(x^3-1)}-\int \ln{(x^3-1)}dx...
23. ### Help whith fourier transform

Think of the cos term and the fact that \cos{\theta}=\frac{e^{i\theta}+e^{-i\theta}}{2} Then using that you should be able to solve the integral. \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{e^{ix^2}+e^{-ix^2}}{2}e^{-ikx}dx
24. ### Help with an integral

Yea I did it and got 0 because you end up integrating \cos(\theta)^3 at the end, which is 0.
25. ### First order differentials

Yea it is non-linear I didn't notice that. What I said wont work.
26. ### First order differentials

It looks like it is first-order homogeneous. Make the substitution z=x/t by dividing by t throughout your t/(2t+x), and then x=tz and x'=z+tz' by the product rule. Now it's in a form you want it for the relevant equations you listed. Just make the substitution the other way when you've solved for z.
27. ### Finding roots in an equation

Have you learned the rational root test?