Need help at Bernoulli Equation

Hi!

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

$lim_{x -> 0^{+}} y(x) <\infty$


The transformation is $u=y^{2}$.

So, $u'+(\frac{1}{x}+1)u=\frac{e^{2x}}{x}$

How can I find a $c=u(1)$ so that $lim_{x -> 0^{+}} u(x) <\infty$ ??

When a differential equation is of the form:

$\dfrac{dy}{dx} + P(x) y = Q(x)$

You can multiply by an integrating factor, $I$, of the form $(*)$ to solve it, where:

$(*) \ I = e^{\int P(x)\ dx}$

Hint:

take L`Hopital`s rule (one) for the limit as x->0 from the right : $(c \times \frac{-e^{-x}}{1})+\frac{2e^{2x}}{3}$

To take L'Hopital's rule, I have to make sure if the denominator and the numerator equal to zero when $x \rightarrow 0^{+}$, don't I???

Correct - l'Hopital won't help you here!

I would separate your solution out into 2 fractions, then substitute the Maclaurin series for the exponentials to see what c has to be to get rid of the terms that are going to cause problems at x = 0.

$u(x)=\frac{e^{-x}}{x}(c+\frac{e^{3x}}{3})$

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

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

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

So, 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 that $c+\frac{1}{3}=0 \Rightarrow c=-\frac{1}{3}$.

Is this correct??