Question

Differential equation - Integrating factor question http://imgur.com/6oiEQ

Answer 1

I've multiplied every term by the factor. Just wondering if I can integrate straight away or is there some step in between that?

Answer 2

I feel like I need simplify somewhere, not sure though.

Answer 3

You should be able to integrate straight away!

Answer 4

But what do I do with the integrating factor that's multiplied by the dy/dt?

Answer 5

e^(x^2/2) y = Int ( x *e^(x^2/2) )dx

Answer 6

u = x^2/2 --> du = xdx

Answer 7

Ok so the integrate of e^(x^2/2) is e^(x^2/2) and the integral of dy/dx is y. I get that. But how did you get the Right hand side?

Answer 8

My problem is that I don't know what do with the e^(x^2/2)*xy

Answer 9

Int ( x *e^(x^2/2) )dx
Let u = x^2/2 --> du = xdx

Answer 10

So \[e ^{u}*xy\]

Answer 11

= Int ( e^u du )

Answer 12

I don't understand :/

Answer 13

The purpose of Integrating factor is to multiply both sides in order to obtain:
e^(x^2/2) * y = Int ( x *e^(x^2/2) ) dx

Answer 14

If I explain in detail, it'd be very lengthy, even more confused!

Answer 15

I understand what the purpose of the integrating factor, its just that I don't where the x from xy is goes to

Answer 16

I think I should have d/dx(e^(x^2/2)x=e^(x^2/2)x

Answer 17

and then integrate.

Answer 18

Well this is what I was planning on doing but wasn't sure.

Answer 19

It's incorrect, because the benefit of multiply Integrating factor is dy term gone!

Answer 20

But do I need to simplify the LHS? the x in xy is completely throwing me

Answer 21

You seems not understand what's integrating factor for:
After multiplying Integrating factor for both sides, the left side turns into ( uv)' form:
which is ( Integrate factor * y )'
Then we Integrate both side :

Answer 22

-> e^(x^2/2) y = Int ( x *e^(x^2/2) )dx

Answer 23

@Chlorophyll

I do understand the LHS is just the opposite of the product rule. the product of e^(x^2/2) is e^(x^2/2) and the product of y is dy/dx.

That's where you got e^(x^2/2) * y from, but the x from xy just dissappears, I don't understand that!!

Answer 24

from the original equation dy/dx+xy=x

Answer 25

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

Answer 26

See, you don't understand the whole left side t urn into product => (uv)'

Answer 27

Is the x derving to be 1?

Answer 28

Are you doing the product rule on e^(x^2/2) and xy letting u=e^(x^2/2) * y and v=xy?

Answer 29

I'm trying to think how to explain easy way!

Answer 30

Sorry for this, you're helping alot.

Answer 31

From the left side, nultiply
e^ ( x^2/2) ( dy/dx + x* y )
e^ ( x^2/2) * dy/dx + e^ ( x^2/2) x * y
= u * y' + u' * y
= ( u * y )'

Answer 32

ahhhhhh makes now after the breakdown!

Answer 33

Finally :D

Answer 34

Now I can integrate by sides cause the d/dx is gone

Answer 35

That's why the left side only have e^ ( x^2/2) * y
Yes, integrate now!

Answer 36

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

Answer 37

+c

Answer 38

The right side:
Int ( x *e^(x^2/2) )dx

Answer 39

= Int ( e^u du )

Answer 40

Did you just take logs of both sides? to cancel out the e on the LHS?

Answer 41

Integral the right side, by u substitution:

Answer 42

I see but, would it not give the same answer if I did, just curious. I mean

x^2/2(y) = LN(1/2x^2*e^(x^2/2)) and then take away e^(x^2/2)+c from both sides to get the equation in terms of y

Answer 43

sorry take away x^2/2

Answer 44

and using my initial values from the question to find out whats the constant

Answer 45

After Int ( e^u du ), the divide both side for e^(x^2/2) to obtain y = ...
Don't forget C

Answer 46

The INV to get particular DE, instead of general one!