Bernoulli Equation

[mathmari]

Hello!!!

I have the following Bernoulli equation:
[tex] 2xyy'+(1+x)y^2=e^{x} [/tex], [tex] x>0 [/tex]

[tex]lim_{x -> 0^{+}} y(x) <\infty [/tex]


The transformation is [tex] u=y^{2} [/tex].

So, [tex] u'+(\frac{1}{x}+1)u=\frac{e^{2x}}{x}[/tex].


How can I find the initial value [tex]u(1)[/tex] so that [tex]lim_{x -> 0^{+}} u(x) <\infty [/tex] ??

 

[Klaas van Aarsen]

Hello!!!

I have the following Bernoulli equation:
[tex] 2xyy'+(1+x)y^2=e^{x} [/tex], [tex] x>0 [/tex]

[tex]lim_{x -> 0^{+}} y(x) <\infty [/tex]


The transformation is [tex] u=y^{2} [/tex].

So, [tex] u'+(\frac{1}{x}+1)u=\frac{e^{2x}}{x}[/tex].


How can I find the initial value [tex]u(1)[/tex] so that [tex]lim_{x -> 0^{+}} u(x) <\infty [/tex] ??

Hey mathmari!

I think that should be:
$$u'+(\frac{1}{x}+1)u=\frac{e^{x}}{x}$$


You can solve this ODE.
Its solution is:
$$u(x)=\frac{A e^{-x} + e^x}{2x}$$
as you can see here.


To make u(x) converge when $x \to 0^+$, you need $A=-1$.
According to l'Hospital, that gives us:
$$\lim_{x \to 0^+} \frac{e^x -e^{-x}}{2x} = \lim_{x \to 0^+} \frac{e^x +e^{-x}}{2} = 1$$
so it really converges.

So:
$$u(1) = \frac{e -e^{-1}}{2}$$

 

[mathmari]

Sorry, I accidentally made a mistake at the exercise..

It's [tex] 2xyy'+(1+x)y^2=e^{2x} [/tex], with solution [tex] u(x)=\frac{3Ae^{-x}+xe^{3x}}{3x} [/tex].

To make u(x) converge when [tex]x\rightarrow 0^{+}[/tex], must [tex]A=0[/tex]??
Then [tex]lim_{x\rightarrow 0^{+}} \frac{xe^{3x}}{3x}=lim_{x\rightarrow 0^{+}} \frac{e^{3x}}{3}=\frac{1}{3}[/tex].
So, [tex] u(1)=\frac{e^{3}}{3}[/tex].
Is this right??

 

[Klaas van Aarsen]

Sorry, I accidentally made a mistake at the exercise..

It's [tex] 2xyy'+(1+x)y^2=e^{2x} [/tex], with solution [tex] u(x)=\frac{3Ae^{-x}+xe^{3x}}{3x} [/tex].

To make u(x) converge when [tex]x\rightarrow 0^{+}[/tex], must [tex]A=0[/tex]??
Then [tex]lim_{x\rightarrow 0^{+}} \frac{xe^{3x}}{3x}=lim_{x\rightarrow 0^{+}} \frac{e^{3x}}{3}=\frac{1}{3}[/tex].
So, [tex] u(1)=\frac{e^{3}}{3}[/tex].
Is this right??

Almost.

I think your solution should be:
$$u(x)=\frac{3Ae^{-x}+xe^{\color{red}{2x}}}{3x}$$

 

[mathmari]

Almost.

I think your solution should be:
$$u(x)=\frac{3Ae^{-x}+xe^{\color{red}{2x}}}{3x}$$

Yes, you're right!!!

I have also thought another way to solve the exercise.. Could you tell me if it's right??


Using the Taylor series for the exponential:

[tex] e^{x}=1+x+\frac{x^2}{2}+... [/tex]

[tex] e^{-x}=1-x+\frac{x^2}{2}-\frac{x^3}{3!}+... [/tex]

[tex] e^{3x}=1+3x+\frac{(3x)^2}{2}+... [/tex]

For small [tex] x>0 [/tex]
[tex] u(x)= \frac{1-x}{x}(c+\frac{1+3x}{3})= (\frac{1}{x}-1)(c+\frac{1+3x}{3})= \frac{1}{x}(c+\frac{1+3x}{3})-(c+\frac{1+3x}{3})[/tex] [tex]= \frac{1}{x}c+\frac{1}{3x}+1- (c+\frac{1+3x}{3})= \frac{1}{x}(c+\frac{1}{3})+1- (c+\frac{1+3x}{3}) [/tex].

Since the term [tex] \frac{1}{x} [/tex] causes problems at [tex] x=0 [/tex], we want to get rid of it, so [tex] c+\frac{1}{3}=0 \Rightarrow c=-\frac{1}{3}[/tex].

But with this way I have found else..

 

[Klaas van Aarsen]

Yes, you're right!!!

I have also thought another way to solve the exercise.. Could you tell me if it's right??


Using the Taylor series for the exponential:

[tex] e^{x}=1+x+\frac{x^2}{2}+... [/tex]

[tex] e^{-x}=1-x+\frac{x^2}{2}-\frac{x^3}{3!}+... [/tex]

[tex] e^{3x}=1+3x+\frac{(3x)^2}{2}+... [/tex]

Aww! You really should become more careful with your exponential powers.
I take it you intended the following?
$$e^{2x}=1+2x+\frac{(2x)^2}{2!}+...$$

For small [tex] x>0 [/tex]
[tex] u(x)= \frac{1-x}{x}(c+\frac{1+3x}{3})[/tex]

Hmm, where did your "c" come from?

Anyway, hold on! We have:
$$u(x) = \frac{3A e^{-x} + x e^{2x}} {3x}$$
Making your substitutions, I get:
$$u(x) =_1 \frac{3A (1-x) + x (1 + 2x)} {3x} = \frac{A (1-x)}{x} + \frac{1+2x}{3}$$
That looks different than what you have (and not only the 2x instead of 3x).
Note that I have used $=_1$ to denote that the equality holds for the first order approximation.
From this, you can again conclude that if the series is to converge, you will at least need that $A=0$.


You have to be careful with series expansions though.
If the resulting series converges, you're okay. Then the original formula will also converge.
But if it diverges, it's inconclusive. A diverging Taylor expansion does not necessarily mean that the original formula is undefined. In this case you're okay, since you're working within the radius of convergence of the exponential expansion.

 

[mathmari]

Ok! Thanks!!

 

[Klaas van Aarsen]

Ok! Thanks!!

Heh. This is an old thread. It actually originates from my birthday.
How come you're getting back to it now?

Edit: Hey! You have an avatar now!!

 

[mathmari]

Heh. This is an old thread. It actually originates from my birthday.
How come you're getting back to it now?

Edit: Hey! You have an avatar now!!

I was looking at my old posts and realized that I hadn't thanked you!