Tough one here!

[Buy2easy.com]

Well it is tough for me any way your guys try this and let me know what you get

y'+((1+x)/x)y=0

it is a differential equation that i have as a hw problem i have the answer(back of the book) but i cant find how to get it?!?

Also some more info it is a linear d.e. use the method for finding linear d.e.'s let me know what you guys get as the answer and if you dont mind let me see the steps you took to get it because i am totally lost in this class!?!?! cal1-3 were a breaze but now there are the hard integrals to solve and i am stuck

OH yea!! No calculators :-) because i can solve it with a TI-89 but i cant seem to get the right solution for the integral over the e

if you do not understand the above statement you probably should not even attempt the problem....the e comes from getting the integral of p(x) over e were p(x) = ((1+x)/x)..and so on

thanks for any help!!!

also if you cant solve it let me know so i do not feel so bad :-)

[bugzpodder]

y'+((1+x)/x)y=0

is in the form of linear differential equation:

dy/dx+p(x)y=q(x)

thus the general solution is given by:

y=(I(u(x)q(x)dx)+C)/u(x)

where u(x)=exp(I(p(x)dx)), aka the integral factor and I denotes the integral sign.

u(x)=e^I(((1+x)/x)dx)=e^I((1/x+1)dx)=e^(I(1/x)dx+I(1)dx)=e^(ln(abs(x))+x+C₀)=abs(x)*e^(x+C₀)

since I(u(x)q(x)dx)=I(0)dx=C₁, where C₁ is some constant

so we have y=C/u(x)

or y=C/(abs(x)*e^(x+C₀))

what year in university do they teach this?

oh and I'm not sure if its right or not, i haven't had any practice with differential equations.

[Buy2easy.com]

i am going into computer & electrical engineering this is my third year last math class i need(i believe) . what is the abs(x)* in your equation mean? the book says that the answer of a linear d.e. is given by y=c/e^ingegral of p(x) or something like that???

also what is the integral of (1+x)/x how i would do it is like the folloowing.....

split it
1/x+x/x ---> 1/x+1
get the integral of that
ln|x|+X
wam bam thank you mam!
but if you plug that into the linear d.e. i gave then you get
y=c/e^ln|x|+x

after the ^ means it is an exponent

but in the back of the book it is nothing like that!

I need some sereous help!!

is my integration wrong or has the teacher done a sucky job teaching me how to do linear d.e.

Thanks!

[bugzpodder]

abs is absolute value. and why don't you tell me what the back of ur book says?? or i can't help u

[bugzpodder]

Originally posted by Buy2easy.com
the book says that the answer of a linear d.e. is given by y=c/e^ingegral of p(x) or something like that???

yea, if the equation is equal to 0

and abs(x) means absolute value of x

so ln(abs(x)) is ln|x|

also what is the integral of (1+x)/x how i would do it is like the folloowing.....

split it
1/x+x/x ---> 1/x+1
get the integral of that
ln|x|+X
wam bam thank you mam!

yes, except i added a C to the end, so i gave:
ln|x|+x+c

but if you plug that into the linear d.e. i gave then you get
y=c/e^ln|x|+x

looks alright, except i would put brackets:

y=c/e^(ln|x|+x+c)=c/(|x|*e^(x+c))

give me the answer in the back of the book

[A$$Bandit]

In the UK, linear differential equations are taught to Further Maths students at 17...

[DavidHooper]

They certainly are - I remember it well . 2nd order DEs were on the OCR syllabus last year as well.

[Destined Soul]

Originally posted by bugzpodder
oh and I'm not sure if its right or not, i haven't had any practice with differential equations.

LOL! You surprise me yet again, BugZ. Have you actually learned that formula before, or do you open a book and start using the equations (minding the limitations, especially in DE)

Destined

[bugzpodder]

nope never done any differential equations and intergration before. but got all i need to solve the problem from: http://www.sosmath.com/

i thought i'd prove this statement from Buy2easy.com wrong (other than showing off of course)

if you do not understand the above statement you probably should not even attempt the problem....

[A$$Bandit]

Which module was that, Mr. Hooper? Although I'm about to start P1, I know most of P1-3 already