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

So I have to use the method of integrating factors. I thought I could do

$${d\over dt}[(1+t^2)y]=2ty+(1+t^2)$$

But that doesn't give me the $4ty$ that I need. Help?

share|cite|improve this question
It's linear. Have you not seen a formula for integrating factor that works for linear equations? – Gerry Myerson Jan 28 '13 at 4:56
Let me suggest you begin to accept answers to your questions. – Did Feb 2 '13 at 10:55

The formula and rational for the integrating factor should be in your text. Use $m=(1+t^2)$ and multiply both sides of the original equation with it. Then write left hand side as a derivative.

Rewrite $a(t)y'+b(t)y=c(t)$ as $y'+(b/a) y= c/a$. Then multiply last equation with $\mu= e^ {\int (b/a) dt}$. Now equation is $\mu y' + (b/a) \mu y = \mu c/a$. Note $\mu'=(b/a) \mu$ so the left hand side of equation is now a complete derivative $(\mu y)'= \mu y' + \mu' y= \mu y' + (b/a) \mu y =\mu c/a$. Next integrate the equation to get $\mu y = \int{\mu c/a dt} +K$ for some constant $K$, and so $y= {{\int{\mu c/a dt} +K} \over \mu}$.

share|cite|improve this answer
That worked. Thanks a lot. – Gamecocks99 Jan 28 '13 at 4:58
Thanks for that comment +1 – Babak S. Jan 29 '13 at 4:44

First of all,


Maybe missing the $y'$ was just a typo.

In any case, first step is to divide by that $y'$ coefficient.


Now we want to find some integrating factor $u$ such that



$$\ln u=2\ln(1+t^2)=\ln[(1+t^2)^2]$$


Note that for finding the integrating factor, the constant of integration isn't important. Had it been included, we would have $u=k(1+t^2)^2$, which would work equally well.

If that seems too complicated, once the left side is in the form $y'+p(t)y$, the integrating factor is simply $e^{\int p(t)dt}$.

share|cite|improve this answer

For every $\color{red}{n}$, $((1+t^2)^\color{red}{n}y)'=(1+t^2)^{\color{red}{n}-1}((1+t^2)y'+2\color{red}{n}ty)$, hence, choosing $\color{red}{n}=2$ and setting $z=(1+t^2)^\color{red}{2}y$, one wants to solve $z'=1/(1+t^2)$, that is, $z=c+\arctan t$, or equivalently, $$ y=\frac{c+\arctan t}{(1+t^2)^2}. $$

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.