Clarifying the Use of Integrating Factors in Exact Differential Equations

[chwala]

Homework Statement

Kindly see the attached....

Relevant Equations

exact differential equations

I am looking at this and i would like some clarity...

at the step where "he let" $μ_y$=0" Could we also use the approach, $μ_x$=0"?
so that we now have,

$μ_y$M=μ($N_{x} -M_{y})$... and so on, is this also correct?

 

[Mark44]

Homework Statement:: Kindly see the attached...
Relevant Equations:: exact differential equations

I am looking at this and i would like some clarity...

View attachment 285140at the step where "he let" $μ_y$=0" Could we also use the approach, $μ_x$=0"?

Since $\mu = \mu(x)$ is a function of x alone, its partial with respect to y must be zero.
No, it's not reasonable to assume that $\mu_x$ would be zero.

 

[chwala]

Since $\mu = \mu(x)$ is a function of x alone, its partial with respect to y must be zero.
No, it's not reasonable to assume that $\mu_x$ would be zero.

i thought its indicated that $μ$ can be a function of $x$ or $y$ only? aaaaargh i have seen the condition...thanks Mark

 

[ergospherical]

at the step where "he let" $μ_y$=0" Could we also use the approach, $μ_x$=0"?
so that we now have,

$μ_y$M=μ(N_{x}$-M_{y})$... and so on, is this also correct?

Yeah, it's okay to try that! There might be an integrating factor $\mu(x)$, or an integrating factor $\tilde{\mu}(y)$, or even both. Or neither.

 

[chwala]

Yeah, it's okay to try that! There might be an integrating factor $\mu(x)$, or an integrating factor $\tilde{\mu}(y)$, or even both. Or neither.

what do you mean by "even both"? i think that's wrong. It can only be one or the other. Not Both! By neither, you are implying that we are not able to find an integrating factor, which of course may be true.

 

[ergospherical]

what do you mean by "even both"? i think that's wrong. It can only be one or the other. Not Both!

Let's see! Let $\omega = y dx - x dy$. Since $\dfrac{1}{N} \left( \dfrac{\partial M}{\partial y}- \dfrac{\partial N}{\partial x}\right) = -\dfrac{2}{x}$ is a function of $x$ only, there's an integrating factor of the form $\mu(x)$. Write\begin{align*}
\dfrac{1}{\mu} \dfrac{d\mu}{dx} &= -\dfrac{2}{x} \\

\mu &= \dfrac{c}{x^2}
\end{align*}It's simplest to take $c=1$ and then you can check it works by noticing that $\mu \omega = d\left( - \dfrac{y}{x} \right)$.

Also, $-\dfrac{1}{M} \left( \dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x}\right) = -\dfrac{2}{y}$ is a function of $y$ only, so there's also an integrating factor of the form $\tilde{\mu}(y)$. Can you work out what it is?

The take away is that integrating factors are not unique.

 

[chwala]

Homework Statement:: Kindly see the attached...
Relevant Equations:: exact differential equations

I am looking at this and i would like some clarity...

View attachment 285140at the step where "he let" $μ_y$=0" Could we also use the approach, $μ_x$=0"?
so that we now have,

$μ_y$M=μ(N_{x}$-M_{y})$... and so on, is this also correct?

this can be used too...dependent on whether we want to end up with a function of $x$ or $y$, which is again dependent on the integrating factor being a 'function of $x$ or$y$'. In this attachment the integrating factor was given as a function of $x$...thus no need to seek another way...

 

[ergospherical]

Generally $\mu(x,y)$ is an integrating factor if $\mu \omega$ is exact, but it's usually quite hard to solve the linear partial differential equation for $\mu$. But there are sometimes special cases where you can find integrating factors depending on one variable only.

 

[chwala]

Let's see! Let $\omega = y dx - x dy$. Since $\dfrac{1}{N} \left( M_y - N_x\right) = -\dfrac{2}{x}$ is a function of $x$ only, there's an integrating factor of the form $\mu(x)$. Write\begin{align*}
\dfrac{1}{\mu} \dfrac{d\mu}{dx} &= -\dfrac{2}{x} \\

\mu &= \dfrac{c}{x^2}
\end{align*}It's simplest to take $c=1$ and then you can check it works by noticing that $\mu \omega = d\left( - \dfrac{y}{x} \right)$.

Also, $\dfrac{-1}{M} \left( M_y - N_x\right) = -\dfrac{2}{y}$ is a function of $y$ only, so there's also an integrating factor of the form $\tilde{\mu}(y)$. Can you work out what it is?

The take away is that integrating factors are not unique.

Generally $\mu(x,y)$ is an integrating factor if $\mu \omega$ is exact, but it's usually quite hard to solve the linear partial differential equation for $\mu$. But there are sometimes special cases where you can find integrating factors depending on one variable only.

but this is a fact no dispute on this...

 

[chwala]

Let's see! Let $\omega = y dx - x dy$. Since $\dfrac{1}{N} \left( \dfrac{\partial M}{\partial y}- \dfrac{\partial N}{\partial x}\right) = -\dfrac{2}{x}$ is a function of $x$ only, there's an integrating factor of the form $\mu(x)$. Write\begin{align*}
\dfrac{1}{\mu} \dfrac{d\mu}{dx} &= -\dfrac{2}{x} \\

\mu &= \dfrac{c}{x^2}
\end{align*}It's simplest to take $c=1$ and then you can check it works by noticing that $\mu \omega = d\left( - \dfrac{y}{x} \right)$.

Also, $-\dfrac{1}{M} \left( \dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x}\right) = -\dfrac{2}{y}$ is a function of $y$ only, so there's also an integrating factor of the form $\tilde{\mu}(y)$. Can you work out what it is?

The take away is that integrating factors are not unique.

i am trying to understand your approach...a bit confusing, this here is clear to me see attachment below:

 

[chwala]

ok i see that they are expressed differently...you brought the negative outside...but in both cases we are either treating the integrating factor solely as a function of $x$ or $y$. Not both.

 

[chwala]

Looking at post $6$, is it necessary to include the constant $c$ when determining the integrating factor when solving this kind of problems?

then letting $c$ = "to a certain value"? unless i am missing something...

i thought the integrating factor $μ(x)$ is simply equal to $\frac {1}{x^2}$...

our brains are thinking differently...$c$ the constant in your post is simply equal to $1$...ok cheers mate