I am not sure how to do this problem. Can you help?
Solve (t^2 + 1)dy/dx = yt-y.
We are asked to find the value of $\displaystyle y$ in $\displaystyle (t^2+1)\frac{dy}{dt}=y(t-1)$.
Like TPH said, this is a separable differential equation so
$\displaystyle
\frac{dy}{y} = \frac{t-1}{t^2+1}dt \Longleftrightarrow \int \frac{dy}{y} = \int \frac{t-1}{t^2+1}dt \Longleftrightarrow \ln{\left |y \right |} = \int \frac{t}{t^2+1}dt - \int \frac{dt}{t^2+1}
$
which gives us
$\displaystyle
\ln{\left |y \right |} = \frac{1}{2}\ln (t^2+1) - \arctan t + C
$
and solving for $\displaystyle y$, we arrive at
$\displaystyle
y=\frac{C\sqrt{t^2+1}}{e^{\arctan t}}
$ for any constant C.
We may verify that this is the general solution by differentiating $\displaystyle y$ and substituting it in the original differential equation.