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Consider the Initial value problem $$y'=f(t)y(t),y(0)=1$$ where $f:\mathbb{R}\rightarrow\mathbb{R}$ is a continuous function. Then the IVP 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 $f.$

The function $f$ is clearly Lipschitz in any bounded domain of $\mathbb{R}^{2}$ but not on whole $\mathbb{R}^{2}$, so i can say that ODE has unique solution on any bounded interval containing $0$ and hence unique solution on whole $\mathbb{R}.$ Am i right in my concept? If i am wrong please suggest me. Thanks in advance.

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    $\begingroup$ Since this is linear time varying ODE of dimension 1 the solution is $y(t)=\exp(\int_0^t{f(s)ds})y(0)$. This solution is unique and globally defined. Thus, as you stated the correct answer is 2. $\endgroup$ – RTJ May 2 '16 at 17:42
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Solutions are indeed unique, but for an arbitrary $f$ the interval may not be the whole $\mathbb R$. From this observation, it follows that the only possibilities would be $2$ and $4$.

In order to find which is the right one, you need to find explicitly some $f$ for which the solution is not defined in $\mathbb R$, or to show that no such $f$ exists.

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  • $\begingroup$ my concept of Lipschitz is not correct??? $\endgroup$ – neelkanth May 2 '16 at 10:17
  • $\begingroup$ Somehow I think that because this equation has special form, the finite time blow-up is not possible here. Of course I might be confusing something, but explicit formula for solution seems to be very nice. $\endgroup$ – Evgeny May 2 '16 at 10:18
  • $\begingroup$ Your concept of Lipschitz is fine, but your conclusion isn't (more precisely: in general it doesn't follow from the Lipschitz property that solutions are global). Of course I agree with @Evgeny, but you asked for a suggestion not for a solution. $\endgroup$ – John B May 2 '16 at 10:23
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Given ODE is $y'-f(t)y=0$ So it is linear ODE with initial condition so by uniqueness theorem has unique solution on $\mathbb{R}$.

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