MHB Solving the Bernoulli Equation: Find Initial Value u(1)

[mathmari]

Hello!

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

lim_{x -> 0^{+}} y(x) <\inftyThe transformation is u=y^{2}.

So, u'+(\frac{1}{x}+1)u=\frac{e^{2x}}{x}.How can I find the initial value u(1) so that lim_{x -> 0^{+}} u(x) <\infty ??

 

[I like Serena]

Hello!

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

lim_{x -> 0^{+}} y(x) <\inftyThe transformation is u=y^{2}.

So, u'+(\frac{1}{x}+1)u=\frac{e^{2x}}{x}.How can I find the initial value u(1) so that lim_{x -> 0^{+}} u(x) <\infty ??

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

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

 

[I like Serena]

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

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

To make u(x) converge when x\rightarrow 0^{+}, must A=0??
Then lim_{x\rightarrow 0^{+}} \frac{xe^{3x}}{3x}=lim_{x\rightarrow 0^{+}} \frac{e^{3x}}{3}=\frac{1}{3}.
So, u(1)=\frac{e^{3}}{3}.
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! :D

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:

e^{x}=1+x+\frac{x^2}{2}+...

e^{-x}=1-x+\frac{x^2}{2}-\frac{x^3}{3!}+...

e^{3x}=1+3x+\frac{(3x)^2}{2}+...

For small x>0
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}) = \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}).

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

But with this way I have found else..

 

[I like Serena]

Yes, you're right! :D

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:

e^{x}=1+x+\frac{x^2}{2}+...

e^{-x}=1-x+\frac{x^2}{2}-\frac{x^3}{3!}+...

e^{3x}=1+3x+\frac{(3x)^2}{2}+...

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 x>0
u(x)= \frac{1-x}{x}(c+\frac{1+3x}{3})

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!

 

[I like Serena]

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! (Mmm)

 

[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! (Mmm)

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