Question

Integrating factor question.

Answer 1

I have th

Answer 2

I'm getting the solution to be y=c

Answer 3

Wolfram says otherwise

Answer 4

I forgot how to do integrating factors, did u multiply both sides by it?

Answer 5

Yes I'll show you my attempt now.

Answer 6

kk I think I'll be able to follow that.

Answer 7

Where did the x go in your solution?

Answer 8

(dy/dx)(e^(x^2/2))+x*e^(x^2/2)*y=x*e^(x^2/2)

Answer 9

I integrated e^(x^2/2)*x

It's equals to e^(x^2/2)+C

Answer 10

\[\frac{\text dy}{\text dx}+xy=x,\qquad x>1,\qquad y(1)=0\]
\[\mu(x)=e^{x^2/2}\]

\[\frac{\text d}{\text dx}\left(y\mu \right)=x\mu\]\[\frac{\text d}{\text dx}\left(y e^{x^2/2} \right)=xe^{x^2/2}\]\[ye^{x^2/2} =\int xe^{x^2/2}\]

Answer 11

Yep

Answer 12

dx

Answer 13

When you integrate the right side, you get the integrating factor plus a constant.

Answer 14

Oh maybe I need you use integration by parts here.

Answer 15

no u use u sub

Answer 16

ok now I finally followed.
u = x^2/2
du = xdx
Integral = (e^u)du
Integral = e^(x^2/2)+c

Answer 17

y(1)=0

Answer 18

Yep

Answer 19

\[ye^{x^2/2} =\int xe^{x^2/2}\text dx\]
let\[x=u\qquad e^{x^2/2}\text dx=\text dv\]\[\text dx=\text du\qquad\qquad \frac{e^{x^2/2}}{x}=v\]

Answer 20

y*e^(x^2/2)=e^(x^2/2)+c
0=e^(1/2)+c
c=-e^(1/2)

Answer 21

y=1-[e^(1/2)]/[e^(x^2/2)]

Answer 22

Unkle why are you using integration by parts?

Answer 23

Well u can use both methods, u sub wld be easier though

Answer 24

Thats I did. However, I solved the ode in wolfram gives a different answer. http://www.wolframalpha.com/input/?i=solve+dy%2Fdx+%2Bxy%3Dx

Answer 25

This is what I have and I cannot see anything wrong with it

Answer 26

No wait it would be y=c/int factor

Answer 27

your right no need for integration by parts,

\[ye^{x^2/2} =e^{x^2/2}+c\]
\[y =1+ce^{-x^2/2}\]

Answer 28

I see what wolfram did, they brought up the int factor and changed the sign.

Answer 29

\[y(x) =1+ce^{-x^2/2}\]
\[y(1) =1+ce^{-1^2/2}=0\]

Answer 30

so c=-e^1/2

Answer 31

\[y(x)=1-e^{(1-x^2)/2}\]

Answer 32

Subtract -1 from both sides.

Answer 33

and divide by the int factor?

Answer 34

to get c on its own.

Answer 35

uhmm did i get it right?

Answer 36

I'm not sure, I'm not getting the same as @amistre64 anyways

Answer 37

i dont recall working in this problem :)

Answer 38

Oh god, I am losing my mind @UnkleRhaukus sorry @amistre64

Answer 39

are you still having trouble with this question @ironictoaster?

Answer 40

I'm not sure how you got the constant from the inital condition

Answer 41

I have y=1+c^-0.5=1

Answer 42

I have y=1+c^-0.5=0

Answer 43

\[y(x) =1+ce^{-x^2/2}\]
sub substituting x=0

\[y(1) =1+ce^{-1^2/2}\]\[0=1+ce^{-1/2}\]\[-1=ce^{-1/2}\]\[-e^{1/2}=c\]

Answer 44

Do you not sub x=1

Answer 45

yeah whops,

Answer 46

So you divided by both sides by the e^...

Answer 47

yeah the last step was to divide by e^-1/2
or to multiply by e^1/2
same thing

Answer 48

I see, my problem was that I did it on a calculator and -1.6....

Answer 49

and got -1.6

Answer 50

\[c=-e^{1/2}\approx-1.65\]

Answer 51

Thats great, thank you.