-2.2.31 First order homogeneous ODE

In summary, the equation x\frac{dv}{dx}+v=1+v+v^2 becomes x\frac{dv}{dx}+v=1+v^2+\arctan(v)=ln|c_1x|. Taking the tangent of both sides gives y/x= tan(ln(|c_1x|)). Multiplying both sides by x gives y= xtan(ln(|c_1x|)).
  • #1
karush
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I
OK going to do #31 if others new OPs
I went over the examples but?
well we can't 6seem to start by a simple separation
I think direction fields can be derived with desmos
 
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  • #2
You want to begin by writing:

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

Next consider the substitution:

\(\displaystyle v=\frac{y}{x}\implies y=vx\implies \frac{dy}{dx}=x\frac{dv}{dx}+v\)

So, make the substitutions and simplify...what do you get?
 
  • #3
Since the first part of the problem is to "show that the given equation is homogeneous" it helps to know what "homogeneous" means here! A differential equation is said to be "homogeneous" if and only if replacing both "x" and "y" by "tx" and "ty", where t is an arbitrary (but non-zero) number, gives exactly the same equation. Of course \(\displaystyle \frac{d(ty)}{d(tx)}= \frac{dy}{dx}\) while replacing x and y in \(\displaystyle \frac{x^2+ xy+ y^2}{x^2}\) by tx and ty gives \(\displaystyle \frac{(tx)^2+ (tx)(ty)+ (ty)^2}{(tx)^2}= \frac{t^2x^2+ t^2xy+ t^2y^2}{t^2x^2}= \frac{t^2(x^2+ xy+ y^2)}{t^2x^2}= \frac{x^2+ xy+ y^2}{x^2}\).

(Notice that both numerator and denominator consist of terms that are quadratic in x and y. That's why we can cancel \(\displaystyle t^2\).)

And the point of that is that we can replace y with y/x and get a simpler equation. Dividing, like MarkFL did, both numerator and denominator on the right by \(\displaystyle x^2\) we get \(\displaystyle \frac{dy}{dx}= \frac{1+ \frac{y}{x}+ \left(\frac{y}{x}\right)^2}{1}= \left(\frac{y}{x}\right)^2+ \frac{y}{x}+ 1\).
Letting \(\displaystyle u= \frac{y}{x}\), \(\displaystyle y= xu\) so \(\displaystyle \frac{dy}{dx}= x\frac{du}{dx}+ u\).
The equation becomes \(\displaystyle x\frac{du}{dx}+ u= u^2+ u+ 1\).
\(\displaystyle x\frac{du}{dx}= u^2+ 1\), a "separable equation".
We can write that as \(\displaystyle \frac{du}{u^2+ 1}= \frac{dx}{x}\) and integrate both sides.
 
  • #4
MarkFL said:
You want to begin by writing:

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

Next consider the substitution:

\(\displaystyle v=\frac{y}{x}\implies y=vx\implies \frac{dy}{dx}=x\frac{dv}{dx}+v\)

So, make the substitutions and simplify...what do you get?
the LHS looks like a product... I think!
 
  • #5
HallsofIvy said:
We can write that as \(\displaystyle \frac{du}{u^2+ 1}= \frac{dx}{x}\) and integrate both sides.
$$\int \dfrac{1}{u^2 +1} du=\int \dfrac{1}{x}dx $$
$$\arctan u = \ln x +c $$
replace u
 
  • #6
$$\int \dfrac{1}{u^2 +1} du=\int \dfrac{1}{x}dx $$
$$\arctan u = \ln x +c $$
replace $u= \frac{y}{x}$

$\arctan \dfrac{y}{x} - \ln x =c$
I think this it thanks everyone gteat help
 
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  • #7
$\arctan \dfrac{y}{x} - \ln x =c$.
I tried plotting this with different values of c but couldn't see any difference
also how do you get "solved" in the title when your done
 
  • #8
MarkFL said:
You want to begin by writing:

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

Next consider the substitution:

\(\displaystyle v=\frac{y}{x}\implies y=vx\implies \frac{dy}{dx}=x\frac{dv}{dx}+v\)

So, make the substitutions and simplify...what do you get?

Next we have:

\(\displaystyle x\frac{dv}{dx}+v=1+v+v^2\)

\(\displaystyle x\frac{dv}{dx}=1+v^2\)

\(\displaystyle \int \frac{1}{v^2+1}\,\frac{dv}{dx}\,dx=\int\frac{1}{x}\,dx\)

\(\displaystyle \arctan(v)=\ln|c_1x|\)

\(\displaystyle v=\tan(\ln|c_1x|)\)

\(\displaystyle y(x)=x\tan(\ln|c_1x|)\)

[DESMOS]{"version":7,"graph":{"viewport":{"xmin":-10,"ymin":-12.5,"xmax":10,"ymax":12.5}},"randomSeed":"1941c0dc1cb008249dbf0302b22e1d8c","expressions":{"list":[{"type":"expression","id":"1","color":"#c74440","latex":"y=x\\ln\\left(\\left|cx\\right|\\right)"},{"type":"expression","id":"2","color":"#2d70b3","latex":"c=0.1","hidden":true,"slider":{"hardMin":true,"hardMax":true,"min":".01","max":"5","step":".001"}},{"type":"expression","id":"3","color":"#388c46"}]}}[/DESMOS]
 
  • #9
Karush, do you see that your solution, $arctan(y/x)- ln(|x|)= c$, is the same as MarkFL's, $y= x tan(ln|c_1x|)$?
(Well, I added the absolute value that should have been in yours. $\int \frac{dx}{x}= ln(|x|)+ C$)

From $arctan(y/x)- ln(|x|)= c$ add ln(|x|) to both sides to get $arctan(y/x)= ln(|x|)+ c$. If we let $c_1= e^c$ (and, of course, $c_1$ is positve) the $c= ln(c_1)$ so $arctan(y/x)= ln(|x|)+ ln(c_1)= ln(|c_1x|)$.

Now, obviously, take the tangent of both sides to get $y/x= tan(ln(|c_1x|)$. Finally multiply both sides by x: $y= xtan(ln(|c_1x|)$.
 
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Related to -2.2.31 First order homogeneous ODE

1. What is a first order homogeneous ODE?

A first order homogeneous ODE (ordinary differential equation) is an equation that involves a single unknown function and its first derivative, where all of the terms in the equation are of the same degree. It is considered homogeneous because the equation can be rewritten in a way that all of the terms have the same units.

2. What does it mean for an ODE to be homogeneous?

A homogeneous ODE is one in which all of the terms in the equation have the same degree, meaning they have the same units. This allows for the equation to be rewritten in a way that is more easily solvable.

3. How do you solve a first order homogeneous ODE?

To solve a first order homogeneous ODE, you can use the method of separation of variables or the method of integrating factors. Both methods involve manipulating the equation to separate the variables and then integrating to find the solution.

4. What is the difference between a homogeneous and non-homogeneous ODE?

The main difference between a homogeneous and non-homogeneous ODE is the presence of a constant term. In a homogeneous ODE, there are no constant terms, while in a non-homogeneous ODE, there is at least one constant term. This makes solving a homogeneous ODE easier, as it can be rewritten in a way that all terms have the same units.

5. What are some real-world applications of first order homogeneous ODEs?

First order homogeneous ODEs have many real-world applications, including in physics, chemistry, and engineering. They are used to model various physical phenomena such as radioactive decay, population growth, and chemical reactions. They are also used in circuit analysis and in the study of fluid dynamics.

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