MHB -2.2.31 First order homogeneous ODE

[karush]

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

 

[MarkFL]

You want to begin by writing:

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

Next consider the substitution:

$$v=\frac{y}{x}\implies y=vx\implies \frac{dy}{dx}=x\frac{dv}{dx}+v$$

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

 

[HallsofIvy]

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 [math]\frac{d(ty)}{d(tx)}= \frac{dy}{dx}[/math] while replacing x and y in [math]\frac{x^2+ xy+ y^2}{x^2}[/math] by tx and ty gives [math]\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}[/math].

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

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 [math]x^2[/math] we get [math]\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[/math].
Letting [math]u= \frac{y}{x}[/math], [math]y= xu[/math] so [math]\frac{dy}{dx}= x\frac{du}{dx}+ u[/math].
The equation becomes [math]x\frac{du}{dx}+ u= u^2+ u+ 1[/math].
[math]x\frac{du}{dx}= u^2+ 1[/math], a "separable equation".
We can write that as [math]\frac{du}{u^2+ 1}= \frac{dx}{x}[/math] and integrate both sides.

 

[karush]

You want to begin by writing:

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

Next consider the substitution:

$$v=\frac{y}{x}\implies y=vx\implies \frac{dy}{dx}=x\frac{dv}{dx}+v$$

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

the LHS looks like a product... I think!

 

[karush]

We can write that as [math]\frac{du}{u^2+ 1}= \frac{dx}{x}[/math] and integrate both sides.

$$\int \dfrac{1}{u^2 +1} du=\int \dfrac{1}{x}dx $$
$$\arctan u = \ln x +c $$
replace u

 

[karush]

$$\int \dfrac{1}{u^2 +1} du=\int \dfrac{1}{x}dx $$
$$\arctan u = \ln x +c $$
replace $u= \frac{y}{x}$

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

 

[karush]

$\arctan \dfrac{y}{x} - \ln x =c$.
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]

You want to begin by writing:

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

Next consider the substitution:

$$v=\frac{y}{x}\implies y=vx\implies \frac{dy}{dx}=x\frac{dv}{dx}+v$$

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

Next we have:

$$x\frac{dv}{dx}+v=1+v+v^2$$

$$x\frac{dv}{dx}=1+v^2$$

$$\int \frac{1}{v^2+1}\,\frac{dv}{dx}\,dx=\int\frac{1}{x}\,dx$$

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

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

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

[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]

 

[HallsofIvy]

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: $y= xtan(ln(|c_1x|)$.