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First order homogeneous ODE

karush

Well-known member
Jan 31, 2012
2,621
Wahiawa, Hawaii
2020_05_14_23.00.48~2.jpgI
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
 

MarkFL

Pessimist Singularitarian
Staff member
Feb 24, 2012
13,666
St. Augustine, FL.
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?
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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.
 

karush

Well-known member
Jan 31, 2012
2,621
Wahiawa, Hawaii
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!
 

karush

Well-known member
Jan 31, 2012
2,621
Wahiawa, Hawaii
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
 

karush

Well-known member
Jan 31, 2012
2,621
Wahiawa, Hawaii
$$\int \dfrac{1}{u^2 +1} du=\int \dfrac{1}{x}dx $$
$$\arctan u = \ln x +c $$
replace $u= \frac{y}{x}$

$\arctan \dfrac{y}{x} - \ln x =c$
I think this it thanks everyone gteat help
 
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karush

Well-known member
Jan 31, 2012
2,621
Wahiawa, Hawaii
$\arctan \dfrac{y}{x} - \ln x =c$.
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
 

MarkFL

Pessimist Singularitarian
Staff member
Feb 24, 2012
13,666
St. Augustine, FL.
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:

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

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

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

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

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

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

 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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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