Homogeneous or not? (First order linear differential equation)

covers911

New member
Joined
Mar 7, 2021
Messages
4
Hi,

I'm wondering about this equation:

y' + 3y = 1

Is this homogeneous or non-homogeneous? And how can I tell? What does homogeneous mean in this context, in straightforward language?

I tried to solve this equation in two ways. First, I actually tried the 'complicated' variation of parameters method, and got the wrong answer:

the CF is: g' + 3g = 0
solving this gives: g = ke^(-3x)
then, after the substitution phase, we end up with: u' = e^(3x)
so: u = (e^3x) / 3 + c (we lose the k here, and I don't know why...)
and: y = ug = 1 / 3 + ce^(-3x)

WRONG.

Second, I tried the simple separation of variables method, and got the right answer (maybe):

rewriting gives: y' / (1 - 3y) = 1
so we end up integrating this expression: 1 / (1-3y) dy
which can be solved using a substitution u = 1 - 3y, which yields: -ln(1 - 3y) / 3 + c

RIGHT - maybe!

Because the answer, acccording to online solvers, is: -ln(|3y - 1|) / 3 + c

Could some kind person clear up my confusion(s) here. Would be most appreciated.

-- C
 

HallsofIvy

Elite Member
Joined
Jan 27, 2012
Messages
7,535
Doesn't you textbook define "homogeneous" for first order differential equations? (Note that the use of "homogenous" for first order equation and higher equations are quite different.)



Any first order differential equation can be written M(x, y)dx+ N(x, y)dy= 0 and is said to be "homogeneous" if M and N are homogeneous (all terms are powers of x and y that add to the same power like "3x^4+ x^2y^2+ xy^3") of the same degree, for example \(\displaystyle x^2dx+ xydy= 0\) is "homogeneous of degree 2"?



But higher order, linear, differential equations are "homogenous" if every term has y or derivative of y term. For example \(\displaystyle y''- 3y'+ 4y= 0\) is "homogeneous" but \(\displaystyle y''- 3y'+ 4y= 3x\) is not.



If what you are really asking is for solutions of y'+ 3y= 1 then, yes, you can write it as \(\displaystyle \frac{dy}{dx}= 1- 3y\) and then \(\displaystyle \frac{dy}{1- 3y}= dx\) and integrating both sides, \(\displaystyle -\frac{1}{3}\ln(1- 3y)= x+ c\).

You dropped the "dx" from the right side of the equation!

(we could divide both sides by -3 to get \(\displaystyle \ln(1- 3y)= -3x+ c\), \(\displaystyle 1- 3y= e^{-3x}e^{c}= Ce^{-3x}\) where \(\displaystyle C= e^c\). Then \(\displaystyle -3y= -1+ Ce^{-3x}\) and \(\displaystyle y= \frac{1}{3}+ c'e^{-3x}\) where \(\displaystyle c'= -C/3\).)



Another way to solve this equation is to first solve the equation y'+ 3y= 0("homogeneous in the sense usually used for higher order equations) which can be written as \(\displaystyle \frac{dy}{dx}= -3y\) and \(\displaystyle \frac{dy}{y}= -3dx/\). Integrating, \(\displaystyle ln(y)= -3x+ C\) so \(\displaystyle y= C'e^{-3x}\). Now we look for a constant solution, y= c, to the entire equation, y'+ 3y= 1 so that y'= 0 and the equation becomes 3c= 1 so c= 1/3 and the entire solution is \(\displaystyle y= C'e^{-3x}+ \frac{1}{3}\) just as before.



As I said before, your mistake is that you completely dropped the "dx"! The solution you say was given by "online solvers", -ln(|3y - 1|) / 3 + c, is also wrong because it is NOT equal to anything! Was the answer possibly x= -ln(|3y - 1|) / 3 + c?
 
Last edited:

nasi112

Full Member
Joined
Aug 23, 2020
Messages
417
Who told you that \(\displaystyle y = \frac{1}{3} + ce^{-3x}\) is WRONG?

It seems to me perfectly correct.

y' + 3y = 1 is non-homogenous

if

y' + 3y = 0, then it is homogenous
 

covers911

New member
Joined
Mar 7, 2021
Messages
4
Any first order differential equation can be written M(x, y)dx+ N(x, y)dy= 0 and is said to be "homogeneous" if M and N are homogeneous (all terms are powers of x and y that add to the same power like "3x^4+ x^2y^2+ xy^3") of the same degree, for example \(\displaystyle x^2dx+ xydy= 0\) is "homogeneous of degree 2"?
Thank you very much for your answer. I do appreciate it. Immediately I have a question regarding the above.

Am I correct to say M(x, y)dx+ N(x, y)dy= 0 is ANY first order differential equation. Using the same symbols, what is a first order _linear_ differential equation?
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
24,653
Thank you very much for your answer. I do appreciate it. Immediately I have a question regarding the above.

Am I correct to say M(x, y)dx+ N(x, y)dy= 0 is ANY first order differential equation. Using the same symbols, what is a first order _linear_ differential equation?
How does your text book answer that question?

Did you do a Google search? Please tell us what you found!
 

covers911

New member
Joined
Mar 7, 2021
Messages
4
Sure, the book says: a.df(x)/dx + b.f(x) + c = 0

Which looks very different. What is this equation with the symbols you use? I am trying to compare the two. All sources seem to provide slightly different information.
 

HallsofIvy

Elite Member
Joined
Jan 27, 2012
Messages
7,535
A linear first order equation is an equation of the for \(\displaystyle f(x)\frac{dy}{dx}+ g(x)y= h(x)\) where f, g, and h can be any functions of x only. We can write that as \(\displaystyle f(x)dy+ (g(x)y- h(x))dx= 0\).

Thinking of that as M(x,y)dy+ N(x,y)dx= 0, we see that M(x,y) must be a function of x only and N(x,y) must be linear in y in order for the equation to be linear.
 

covers911

New member
Joined
Mar 7, 2021
Messages
4
I thought N must be a function of x only too (for the equation to be linear)?

PS: Also, you seem so casual about manipulating the dx and dy, it's intimidating!
 
Top