Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
1  
ahh so you mean that the question asks for the solution y=y(t) defined on (−∞,−1) and (−1,+∞) that satisfies the given equation? –  Tengu Mar 11 '13 at 15:02
    
Yes, it should be so. –  Sami Ben Romdhane Mar 11 '13 at 15:07
    
I see. Then it makes sense. Thank you very much! –  Tengu Mar 11 '13 at 15:15
    
You're welcome. –  Sami Ben Romdhane Mar 11 '13 at 15:16
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.