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.

Consider the initial value problem $$y’(t)=f(t)y(t), \;y(0)=1$$ where $f:\mathbb{R}\to\mathbb{R}$ is continuous. Then this initial value problem has:

  1. Infinitely many solutions for some $f$.
  2. A unique solution in $\mathbb{R}$.
  3. No solution in $\mathbb{R}$ for some$ f$.
  4. A solution in an interval containing $0$, but not on $\mathbb{R}$ for some $f$.

Can anyone help me finding which of the options are correct? Thanks.

share|improve this question
    
Maybe $\frac{\mathrm d t}{f} = f(t) \mathrm d t$? –  FrenzY DT. Dec 6 '12 at 11:47
2  
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Please consider rewriting your post. –  Did Dec 6 '12 at 11:49

2 Answers 2

Hint: Try to find a function $G$ such that $z:t\mapsto\mathrm e^{G(t)}y(t)$ is such that $z'(t)=0$ for every $t$.

share|improve this answer

$$\frac{dy(t)}{y(t)} = f(t)dt$$

$$\implies \log y(t) = \int_0^tf(t)dt +A$$ now use $y(0) =1$

share|improve this answer
    
after solving this i get $\implies \log y(t) = \int_0^tf(t)dt $.then what would be the conclusion?i believe (b) would be the answer then. am i right? –  digu Dec 6 '12 at 12:05

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.