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Determining if equation is linear in indicated dependent variable

find_the_fun

Active member
Feb 1, 2012
166
I was stuck on a question and went back to look at an example but couldnt understand the example. It has \(\displaystyle (y-x)dx +4xy dy = 0\) is linear in y because it can be written as \(\displaystyle 4xy'+y=x\)

First of all what are dx and dy doing in the equation? They look to be from the end of an integral. Second, how did the equatin get rewritten like that?
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: determining if equation is linear in indicated dependent variable

Did you intend the original expression to instead be the equation:

\(\displaystyle 4x\,dy+(y-x)\,dx=0\) ?

This is in differential form. And yes, integrals also contain a differential (after the integrand usually), to denote with respect to what variable we are integrating.
 

find_the_fun

Active member
Feb 1, 2012
166
Re: determining if equation is linear in indicated dependent variable

Did you intend the original expression to instead be the equation:

\(\displaystyle 4x\,dy+(y-x)\,dx=0\) ?

This is in differential form. And yes, integrals also contain a differential (after the integrand usually), to denote with respect to what variable we are integrating.
No I didn't inted for the original expression to be that.
So the statment could be rewritten as \(\displaystyle \frac{d}{dx}(y-x)+\frac{d}{dy}4xy=0\) ?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: determining if equation is linear in indicated dependent variable

We cannot take an expression and convert it into an equation.
 

find_the_fun

Active member
Feb 1, 2012
166
Re: determining if equation is linear in indicated dependent variable

We cannot take an expression and convert it into an equation.
I see what you mean now and yes you were right. Sorry about that. I thought your point was to move terms around and didn't look at the =0 at the end. I should have been more careful.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: determining if equation is linear in indicated dependent variable

There is still a problem, since:

\(\displaystyle (y-x)\,dx +4xy\,dy=0\)

cannot be rewritten as:

\(\displaystyle 4xy'+y=x\)

or in standard linear form:

\(\displaystyle y'+\frac{1}{4x}y=\frac{1}{4}\)

However, the equation:

\(\displaystyle (y-x)\,dx+4x\,dy=0\)

can.
 

find_the_fun

Active member
Feb 1, 2012
166
Re: determining if equation is linear in indicated dependent variable

DSC_0232.jpg
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: determining if equation is linear in indicated dependent variable

Unfortunately, the book contains a typo.
 

find_the_fun

Active member
Feb 1, 2012
166
Re: determining if equation is linear in indicated dependent variable

Ok well the original question I was working on is
Determine whether the given equation is linear in the indicated dependent variable by matching it with the first differential equation given in (7).

\(\displaystyle u dv+(v+uv-ue^u)du=0\) in v; in u​

I'm not sure if I'm doing this right. So \(\displaystyle u dv=0\) since u is treated like a constant. \(\displaystyle (v+uv-ue^u)du=v-ue^u-e^u\) since v is treated like a constant. Am I supposed to be looking at the dv and du to know which variable to differentiate with respect to? How does the "in v; in u" make a difference?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: determining if equation is linear in indicated dependent variable

Okay, we are given the ODE:

\(\displaystyle u\,dv+(v+uv-ue^u)\,du=0\)

Without knowing the form given in (7), we may still assert:

If it is linear in $v$, then we will be able to rewrite the ODE in the form:

\(\displaystyle \frac{dv}{du}+f(u)v=g(u)\)

and if it is linear in $u$, then we will be able to rewrite the ODE in the form:

\(\displaystyle \frac{du}{dv}+f(v)u=g(v)\)
 

find_the_fun

Active member
Feb 1, 2012
166
Re: determining if equation is linear in indicated dependent variable

Okay, we are given the ODE:

If it is linear in $v$, then we will be able to rewrite the ODE in the form:

\(\displaystyle \frac{dv}{du}+f(u)v=g(u)\)
This is not apparent to me. This is the first time I've seen an equation with du or dv just floating around.
 
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: determining if equation is linear in indicated dependent variable

As an example consider the equation:

\(\displaystyle y=x^2\)

Differentiating with respect to $x$, we may write the ODE:

\(\displaystyle \frac{dy}{dx}=2x\)

In differential form, this ODE is:

\(\displaystyle dy=2x\,dx\)