Determining if equation is linear in indicated dependent variable

[find_the_fun]

I was stuck on a question and went back to look at an example but couldn't 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?

 

[MarkFL]

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]

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 statement could be rewritten as \(\displaystyle \frac{d}{dx}(y-x)+\frac{d}{dy}4xy=0\) ?

 

[MarkFL]

Re: determining if equation is linear in indicated dependent variable

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

 

[find_the_fun]

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]

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]

Re: determining if equation is linear in indicated dependent variable

View attachment 1275

 

[MarkFL]

Re: determining if equation is linear in indicated dependent variable

Unfortunately, the book contains a typo.

 

[find_the_fun]

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]

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]

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.

 

[MarkFL]

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:

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