Am i using the wrong method? not seperable?

[mr_coffee]

Hello everyone, Another diff EQ problem, he gave us like 100. I can't seem to figure what method to use on this one. I tried seperable but that isn't working. Here it is:
Find the particular solution of the differential equation
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/f3/6548289ff6de84dd65caa37d658bfa1.png
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/b1/8df613430b792391dbdebef80a9c361.png
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/b3/33fc13f666a02da7144a5fb4c3dd661.png = ?

dy = (7*cos(x)-y*cos(x))dx;
theres no way i can get it so dx is on 1 side and dy on the other.
if its using integrating factor I'm also stuck because what would be p(t)?
y' + ycos(x) = 7cos(x);
would i let p(t) = y? then
solve using that method? But i always thought it had to be a function of x? so would it be
p(t) = cos(x)?
then do i integrate e^(cos(x)) and get e^(sin(x)) as the integrating factor?

 

[StatusX]

Don't worry about whether it's p(t) or p(x), those are dummy variables. You should know that by now. But yes, you do use an integrating factor, and you're on the right track with e^sin(x). Try multiplying across by it.

 

[HallsofIvy]

Hello everyone, Another diff EQ problem, he gave us like 100. I can't seem to figure what method to use on this one. I tried seperable but that isn't working. Here it is:
Find the particular solution of the differential equation
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/f3/6548289ff6de84dd65caa37d658bfa1.png
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/b1/8df613430b792391dbdebef80a9c361.png
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/b3/33fc13f666a02da7144a5fb4c3dd661.png = ?

dy = (7*cos(x)-y*cos(x))dx;
theres no way i can get it so dx is on 1 side and dy on the other.

What do you mean? You have dx on one side and dy on the other!
If you mean you can't get only y on one side and x on the other that's wrong too: dy= (7- y)cos(x)dx so
[tex]\frac{dy}{7- y}= cos(x)dx[/itex]
That's easy to integrate. -ln|7- y|= sin(x)+ C. Taking the exponential of both sides, the GENERAL solution is
$$y= Ce^{-sin x}$$

if its using integrating factor I'm also stuck because what would be p(t)?
y' + ycos(x) = 7cos(x);
would i let p(t) = y?

Try thinking instead of just memorizing formulas! Whatever general formula you were given p(t) makes no sense because there is no "t" in the problem to begin with. y is the dependent variable because there is a y'- which means dy/dx. x is the independent variable. The coefficient of y is cos(x).

then
solve using that method? But i always thought it had to be a function of x? so would it be
p(t) = cos(x)?
then do i integrate e^(cos(x)) and get e^(sin(x)) as the integrating factor?

NO, the integral of e^cos(x) is NOT e^sin(x). Why would you think such a thing?

 

[mr_coffee]

Somtimes I wonder how i got to Differential equations or how I get A's in all my math classes, sure i suck in a big way on hw, but i think i get my **** in gear when exam time comes around hah.

Ivey, i tred ur general solution and plugged in the I.C. and got:
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/d9/7a7b29742173f399ca1f76418178681.png
but it said it was incorrect :(

 

[StatusX]

I think HallsofIvy meant Ce^{-sin x}+7.

 

[mr_coffee]

GOAL! thanks Status and Ivey, i'll ask my mum to cook up some cookies and i'll e-mail them to you guys. :D

 

[HallsofIvy]

Ouch! I forgot the 7!