1. ## Differential Equations

I am not sure how to do this problem. Can you help?

Solve (t^2 + 1)dy/dx = yt-y.

2. Is t a constant or did you mean $\displaystyle \frac{dy}{dt}$?

3. Originally Posted by graceofayak
I am not sure how to do this problem. Can you help?

Solve (t^2 + 1)dy/dx = yt-y.
$\displaystyle (t^2+1)y' = y(t-1)$

$\displaystyle \frac{y'}{y} = \frac{t-1}{t^2+1}$ or $\displaystyle y=0$

This is seperable. Solve.

4. ## Diff. equation

dy
dt

5. 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.

6. Thank you all for your help!!!!