Given the differential equation $(1+t^2)dy/dt+4ty=(1+t^2)^{-2}$ 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? 
 A: 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 $y= {{\int{\mu c/a dt} +K} \over \mu}$.
A: 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 $e^{\int p(t)dt}$.
A: 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, 
$$
y=\frac{c+\arctan t}{(1+t^2)^2}.
$$
