# Solution to a first order differential equation

I am solving a differential equation $\displaystyle \frac {dy}{dt} =\frac {y+1}{t+1}$.

I got the solution $y=c(t+1)-1$, $c$ a constant.

But the handout by my professor says

"The solution is $y=c(t+1)-1$, $c$ a constant. where $t \neq -1$. But it does not mean that $y$ is not defined at $t \neq -1$. It means that $y$ can take any value at $t=-1$ as long as it satisfies $y=c(t+1)-1$."

It does not make much sense to me because if $y$ satisfies the equation $y=c(t+1)-1$ at $t=-1$, then the solution will be just $y=c(t+1)-1$, $c$ a constant.

And I believe my solution indeed makes sense because $y=c(t+1)-1$, $c$ a constant is defined on $\mathbb R$ and differentiable, $y'=c$ for any $t \in \mathbb R$. On the other hand, plugging this in to $\displaystyle \frac {y+1}{t+1}$ gives me $c$ for any $t \in \mathbb R$.

Would you explain what the handout is saying?

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What happens to $\displaystyle \frac {dy}{dt} =\frac {y+1}{t+1}$ at $t = -1$? – Amzoti Mar 11 '13 at 14:53
Try to find a particular solution at $t=-1$. – Mhenni Benghorbal Mar 11 '13 at 14:56
Check the existence and uniqueness theorem. – Mhenni Benghorbal Mar 11 '13 at 15:05

Since the derivative $\frac{dy}{dt}$ is not defined at $t=-1$ then all that can be found is solutions of the differential equation on the intervals $(-\infty,-1)$ and $(-1,+\infty)$ separately.