[SOLVED] Ingegrating factors

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February 3rd 2009, 04:55 PM
ronaldo_07

[SOLVED] Ingegrating factors

Assign to each of the following differential equations the integrating factor/s (if there are any listed, multiple answers possible).

A) (1 + y^2)/x^2 + y′ = 0 I) μ = 1/(1 + y^2)

B) y/(4 + x) + y′ = 0 II) μ = 1/y

C) 1 + (1 + x)yy′ = 0 III) μ = 4 + x

A= I
B= III
C= II
February 4th 2009, 01:43 PM
ronaldo_07

Quote:

Originally Posted by ronaldo_07

A= I
B= III
C= II

Can anyone confirm if this is correct?
February 4th 2009, 03:10 PM
Danny

Quote:

Originally Posted by ronaldo_07

Assign to each of the following differential equations the integrating factor/s (if there are any listed, multiple answers possible).

A) (1 + y^2)/x^2 + y′ = 0 I) μ = 1/(1 + y^2)

B) y/(4 + x) + y′ = 0 II) μ = 1/y

C) 1 + (1 + x)yy′ = 0 III) μ = 4 + x

A= I
B= III
C= II

Check on what you post in (A) - it's not linear
(B) is correct and check on your post on (C) because it has an integrating factor but not one that is list here.
February 4th 2009, 03:14 PM
ronaldo_07

A- If its not linear then there are no integrating factors?
February 4th 2009, 03:19 PM
Danny

Quote:

Originally Posted by ronaldo_07

A- If its not linear then there are no integrating factors?

Good point. Usually integrating factor are used for linear ODE but if you write your equation

$\frac{1+y^2}{x} \; dx + dy = 0$

then ask for $\mu$ such that your equation becomes

$\mu \frac{1+y^2}{x} \; dx + \mu\; dy = dF = 0$ , then yes, you are correct.
February 8th 2009, 10:55 AM
1234567

What would be the integrating factor for for C, if it is not in the list?
February 8th 2009, 11:02 AM
Danny

Quote:

Originally Posted by 1234567

What would be the integrating factor for for C, if it is not in the list?

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

if you write the equation as

$dx + (1 + x)y dy = 0$

then it would be

$\mu = \frac{1}{1+x}$
February 8th 2009, 11:05 AM
1234567

Quote:

Originally Posted by danny arrigo

c) $1 + (1 + x)yy′ = 0$

if you write the equation as

$dx + (1 + x)y dy = 0$

then it would be

$\mu = \frac{1}{1+x}$

Thak for that, was not really sure what it could have been

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