Question

1st order diff eq.: (dy)/(dx) - y/x = xe^x ; y(1) = e - 1

Answer 1

cool linear DE

Answer 2

@sami-21 your turn

Answer 3

.....yaaaay..... jerkface... jk

Answer 4

your in for a lot more integration :p >:)))

Answer 5

mew!!!! or mu!

Answer 6

this time you do 3 integration >:)))

Answer 7

yup! looking forward to learning it

Answer 8

....if you're lucky

Answer 9

....noooo, i should be finding an integrating factor,.....

Answer 10

sami's taking a long time to type something...

Answer 11

must be a latex bomb

Answer 12

.....life story?

Answer 13

could be

Answer 14

ok .
first find the integrating factor.
\[\Large e^{\int\limits_{}^{}\frac{-1}{x}dx} = e^{-\ln(x)}=e^{\ln(x)^{-1}}=x^{-1}=\frac{1}{x}\]

Answer 15

multiply the whole equation with integrating factor.

Answer 16

wait, how did you get the first part?

Answer 17

are you familiar with linear equation ?

Answer 18

if shown an example, yes

Answer 19

i'm a visual guy

Answer 20

oh!

Answer 21

ok look this at site
http://www.sosmath.com/diffeq/first/lineareq/lineareq.html

Answer 22

P(x)y = y/x = (1/x)y :::: mew(x) = e^(integral: dx/x)

Answer 23

yes it is -(1/x y) so e^integral (-1/x)

Answer 24

forgot the negative, thank you

Answer 25

ok, so now i multiply that to the whole equation?

Answer 26

im, trying to do that math, but i am having no luck

Answer 27

d/(dx)[e^(-Ln|x|)y] = xe^x * e^-ln|x| ???? whaaat?

Answer 28

hmm
its not that hard :P
just multiply this and you will have
1/x(dy/dx)-y/x^2=e^x
is it ok now ??

Answer 29

now how did the right side come out to only e^x?

Answer 30

x*1/x= ????????????????????

Answer 31

and the step before that?

Answer 32

which one ????

Answer 33

why are you dividing it by x?

Answer 34

i said multiply the whole equation with \[\frac{1}{x}\]

Answer 35

ooooooh, i see now, e^-lnx = 1/x......

Answer 36

ok, so how do i proceed from 1/x(dy/dx)-y/x^2=e^x ?

Answer 37

put it all on the right side except for dy/dx and then solve for y?

Answer 38

ok
you can see the left the side is product rule of differentiation for
d/dx(1/x*y)

Answer 39

i did not see this...

Answer 40

differentiate d/dx(1/x*y) use product rule.
\[\Large \frac{d}{dx}(\frac{1}{x}y)=\frac{1}{x}\frac{dy}{dx}-\frac{1}{x^2}*y\]
is it ok now ???????

Answer 41

yes

Answer 42

then the = e^x?

Answer 43

i was just trying to show you the left side
so it becomes
\[\Large \frac{d}{dx}(\frac{1}{x}y)=e^x\]
it it ok ??

Answer 44

and then I integrate?

Answer 45

or get y by itself again?

Answer 46

yes integrate both sides .
the left side derivative and integral will cancel each others.
you will have
\[\Large \frac{1}{x}*y=\int\limits_{}^{}e^x\]
i hope you can do this from here :)))))))

Answer 47

it wouldnt be easier to get y by itself?

Answer 48

but would I get -2x^(-2) * (1/2)y^2 = e^x?

Answer 49

or would i use the product rule again?

Answer 50

since you got
\[\Large \frac{1}{x}*y=\int\limits_{}^{}e^x\]
just you need to integrate right side only
\[\Large \frac{1}{x}y=e^x+C\]
or by multiplying x
\[\Large y=xe^x+Cx\]

Answer 51

oye! so, after you multiply mew back in, how did (1/x)(dy/dx)−(y/x^2) = e^x become (dy/dx)((1/x)*y) = e^x

Answer 52

oh, okay......

Answer 53

because it was the result of product rule so i wrote (dy/dx)((1/x)*y) = e^x
the left side as

Answer 54

so then e-1 = e^x + c?

Answer 55

you do not know integration of e^x ???

Answer 56

i thought the integral of e^x WAS e^x...

Answer 57

yes it is .

Answer 58

i was plugging in the initial value

Answer 59

ok
plug in in the
y=xe^x+Cx
e^-1=e+C
C=e^-1-e
Or
C=1/e-e
Or
C=(1-e^2)/e

Answer 60

oh, ok

Answer 61

thank you

Answer 62

Thanks :P

Answer 63

woah this is still ongoing? :o

Answer 64

several breaks