Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Im solving initial value problem $$ \frac {dy}{dx} + xy = xy^2; y(0)=3$$ After applying Bernoulli's equation method i obtained $$ \frac {dv}{dx} -xv = -x$$ So, $$ p(x) = -x, q(x) = -x $$ For finding integrating factor $$u(x)=e^ {-\int xdx}=e^ {-\frac {x^2}{2}}$$ $$ y= \frac {\int u(x)q(x)dx+C}{u(x)}={\int u(x)q(x)dx}$$ So, $$-\int xe^ {-\frac {x^2}{2}}dx $$ Please help further or guide me if i did something wrong.

share|cite|improve this question
$\int u(x) q(x) dx \ne - \int e^{-x^2/2} dx $. – Hans Engler Nov 23 '12 at 15:37
Sorry, I didn't get it, how? – TPSstar Nov 23 '12 at 15:38
$q(x) = -x$, not $q(x) = -1$. Check all your steps. The integral can then be found in closed form. – Hans Engler Nov 23 '12 at 16:15
Thanks i didn't notice that :) – TPSstar Nov 23 '12 at 16:51
up vote 1 down vote accepted

This seems fine. the only point is $v=\frac1y$, so $y=\frac1v$ and $$u(x)v(x)-1\cdot\frac13=u(x)v(x)-u(0)v(0)=\int_0^x (u(t)v(t))'=\int_0^x u(t)q(t)dt=-\int_0^x te^{-\frac{t^2}{2}}dt$$ So $$v(x)=e^{\frac{x^2}{2}}\left(\frac13-\int_0^x te^{-\frac{t^2}{2}}dt\right),\hspace{10pt} y(x)=\frac{1}{v(x)}$$

share|cite|improve this answer
How do we plug y(0)=3 in equation? – TPSstar Nov 23 '12 at 15:32
Since $v=\frac1y$, we have $v(0)=\frac13$ and I used it at the very beginning of the first line. – Dennis Gulko Nov 23 '12 at 15:37
Thank you got it :) – TPSstar Nov 23 '12 at 15:39

Why don't you try separation of variables? This leads to a closed form solution that exists for all $x < \sqrt{2 \log \frac{3}{2}}$.

share|cite|improve this answer
In that case there is something wrong. If there is a closed form for $y(x)$, then we can find a closed form for $\int_0^x e^{-\frac{t^2}{2}}dt$, which does not exist. – Dennis Gulko Nov 23 '12 at 15:39
there is an error in the original post, see my comments. – Hans Engler Nov 23 '12 at 16:14

$$ \mbox{Set}\ {\rm y}\left(x\right) \equiv {\rm f}\left(x^{2}\right)\equiv{\rm F}\left(x\right) $$

$$ \mbox{You get}\quad {\rm F}'\left(x\right) ={{\rm F}^{2}\left(x\right) - {\rm F}\left(x\right)\over 2} $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.