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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.

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Maybe $\frac{\mathrm d t}{f} = f(t) \mathrm d t$? –  FrenzY DT. Dec 6 '12 at 11:47
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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 $t$.

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

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

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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

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