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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?

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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
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3 Answers

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 $y= {{\int{\mu c/a dt} +K} \over \mu}$.

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That worked. Thanks a lot. –  Gamecocks99 Jan 28 '13 at 4:58
    
Thanks for that comment +1 –  B. S. Jan 29 '13 at 4:44
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First of all,

$$[(1+t^2)y]'=(1+t^2)y'+2ty$$

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

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

$$y'+\frac{4ty}{1+t^2}=(1+t^2)^{-3}$$

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

$$u(y'+\frac{4ty}{1+t^2})=(uy)'=uy'+u'y$$

$$u'=\frac{4tu}{1+t^2},\frac{u'}u=\frac{4t}{1+t^2}$$

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

$$u=(1+t^2)^2$$

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 $e^{\int p(t)dt}$.

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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, $$ y=\frac{c+\arctan t}{(1+t^2)^2}. $$

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