Since you guys are so smart...

Discus: Individual Schools: US News Top 25: Massachusetts Institute of Technology: 2004 Archive: Since you guys are so smart...
 By Seventoedsloth (Seventoedsloth) on Wednesday, March 17, 2004 - 04:59 pm: Edit

why not help me out with my differential equations review. Seriously I need some help.
y'=xe^(-sinx)-ycos(x)
and
(3y^2+2y)y' = xcos(x)

If you guys can help any that would be awesome

 By Amylase (Amylase) on Wednesday, March 17, 2004 - 05:17 pm: Edit

post it again in the SAT/ACT test prep thread, more test and math genius there.

this thread is filled with jubilant accepted and depressed rejected kids like me,

 By Whoami (Whoami) on Thursday, March 18, 2004 - 09:56 pm: Edit

(Using cal 2 knowledge)

As for the first equation, it is a linear first-order differential equation.

Whenever an equation is of the form:
y' + P(x)*y=Q(x), the solution is
y = 1/v(x) * int(v(x)Q(x)dx) where v(x)=e^int(P(x)dx)

the proof is lengthy, read the cal or diff eq book if you are interested.

anyway, i think the solution to the first equation is :

1/(e^sin(x)) * x^2/2

As for the second equation:
it is a first-order separable differential equation.

In fact, it's already seperated. just multiply both sides of the equation by dx and integrate both sides. you'll need to integrate the right hand side by parts. you should get at the end:

y^3+y^2=x*sin(x)+cos(x)+C

where C is the combined constant of integration for both integrals
---------------------------

MIT Rejection pride (cuz MIT dont need anymore smart ppl)

 By Rono_G (Rono_G) on Saturday, March 20, 2004 - 06:50 am: Edit

man, i am not going to MIT... did you hear that Babson kids even made a song about how geeky MIT kids can get... man, no way! never never never... and plus, Jim Bronsky of MIT - I don't know how he got in (last year) with an SATII score of 1350 (all three scores combined).

 By Stevennle (Stevennle) on Saturday, March 20, 2004 - 12:59 pm: Edit

y'=xe^(-sinx)-ycos(x)

It's a first-order linear differential equation.

dy/dx + ycos(x) = xe^(-sin(x))

So you need to find the integrating factor.
Which is: e^(sin(x))

d/dx[ e^(sin(x))y ] = x

Integrate both sides

e^(sin(x))y = 1/2x^2 + c
y = 1/2e^(-sin(x))x^2 + ce^(-sin(x))

 By Whoami (Whoami) on Saturday, March 20, 2004 - 03:26 pm: Edit

hmm i thought the solution to a first order linear diff eq was

Whenever an equation is of the form:
y' + P(x)*y=Q(x), the solution is
y = 1/v(x) * int(v(x)Q(x)dx) where v(x)=e^int(P(x)dx)

where v(x) is the integrating factor

 By Candi1657 (Candi1657) on Saturday, March 20, 2004 - 05:09 pm: Edit

This is Calc 1.

 By Matthias (Matthias) on Saturday, March 20, 2004 - 08:42 pm: Edit

I learned first order differential equations in Calc II (key word: learned, I forgot all of that jib-jab-gibberish a few weeks later). Yee-haw.

 By Nhlgoalie (Nhlgoalie) on Sunday, March 21, 2004 - 11:25 am: Edit

Correct Solution to First Order Linear Diff Eq:

Whenever an equation is of the form:
y' + P(x)*y=Q(x), the solution is
y = 1/v(x) * int(v(x)Q(x)dx) where v(x)=e^int(P(x)dx)

For sure man...

Someone in here who's at MIT (or some USAMO person or something) should type up some nice differential eqs. for us all to destroy