# Thread: non-exact integrating factors vs homogeneous first order equations [example problem]

1. ## non-exact integrating factors vs homogeneous first order equations [example problem]

the problem is : (x+y) dx - x dy = 0

Let M = x+y , N = -x

my first attempt at the solution was via non-exact equations and integrating factors; i noticed that

M_y = 1 and N_x = -1 ; therefore i created an integrating factor mu(x) = (M_y-N_x)/ N
which is equal to -2/x.

therefore my integrating factor 'should' be exp[integral(-2/x) dx] = x^-2

then M* = M*mu(x) = 1/x + y/x^2 & N* = -1/x
integrating these both with respect to x and y respectively gives :
int [M* dx] = ln x - (1/3)* y/x^3 + g(y)
int [ N* dy] = -y/x + h(x)

which i could not succeed in matching to find a family of equations...

is there something wrong with my proccess or am i not allowed to use the method of integrating factors with nonexact equations on homogenous ODEs? my textbook supplies a simple solution using the fact that both M and N are both of degree 1, i had missed this when solving the problem and took the long way--but im not sure why it doesnt seem to work.

2. Originally Posted by damnruskie
the problem is : (x+y) dx - x dy = 0

Let M = x+y , N = -x

my first attempt at the solution was via non-exact equations and integrating factors; i noticed that

M_y = 1 and N_x = -1 ; therefore i created an integrating factor mu(x) = (M_y-N_x)/ N
which is equal to -2/x.

therefore my integrating factor 'should' be exp[integral(-2/x) dx] = x^-2

then M* = M*mu(x) = 1/x + y/x^2 & N* = -1/x
integrating these both with respect to x and y respectively gives :
int [M* dx] = ln x - (1/3)* y/x^3 + g(y)
int [ N* dy] = -y/x + h(x)

which i could not succeed in matching to find a family of equations...

is there something wrong with my proccess or am i not allowed to use the method of integrating factors with nonexact equations on homogenous ODEs? my textbook supplies a simple solution using the fact that both M and N are both of degree 1, i had missed this when solving the problem and took the long way--but im not sure why it doesnt seem to work.
There are a few ways to solve this and what you tried should work.

Your integrating factor is correct this gives the equation

$\displaystyle \left( \frac{1}{x}+\frac{y}{x^2}\right)dx-\frac{1}{x}dy=0$

This gives that

$\displaystyle \frac{\partial f}{\partial y}=-\frac{1}{x} \implies f(x,y)=-\frac{y}{x}+g(x)$

Now taking the partial with respect to x gives

$\displaystyle \frac{\partial f}{\partial x}=\frac{y}{x^2}+g'(x)=\frac{1}{x}+\frac{y^2}{x} \implies g'(x)=\frac{1}{x} \implies g(x)=\ln|x|$

So the final solution is

$-\frac{y}{x}+\ln|x|=c$

Note this equation can also be put in the form

$\displaystyle \frac{dy}{dx}-\frac{1}{x}y=1 \iff \frac{d}{dx}\left[ \frac{1}{x}y\right]=\frac{1}{x} \implies \frac{y}{x}=\ln|x|+c \iff y=x\ln|x|+cx$

3. oh wow, how embarrassing, i found my error - a simple integration miscalculation stemming from a lack of coffee so late at night, apologies and thanks for the help on this example.

As a relatively related aside -> will there every be scenarios in which the method of integrating factors would be unable to solve first order ( or any order) linear homogeneous equations? intuitively it feels to me as if the process for r h.o.d.es [using the substitution y=xv and making a separable equation in x & v] is a sort of sub-case of the integrating factor method... i cant really think of any counterexamples, though my differential equations skills are still weak.

,

,

,

,

,

,

,

,

,

,

,

,

,

,

# non exact differential equationsexample

Click on a term to search for related topics.