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I came across this problem which says: Consider the equation: $$\frac{dy}{dt}=(1+f^{2}(t))y(t); \quad t\geqslant 0,$$ where $f$ is bounded continuous function on $[0,\infty)$.Then which of the following options is correct?

(a) This equation admits a unique solution $y(t)$ and further $\lim_{t\to\infty}y(t)$ exists and finite ,

(b) This equation admits 2 linearly independent solutions,

(c) This equation admits a bounded solution for which $\lim_{t\to\infty}y(t)$ does not exist,

(d) this equation admits a unique solution and further, $\lim_{t\to\infty}y(t)=\infty$.

I have taken $f(t)$ to be $1/(1+t)$ so that $f$ is bounded and continuous in the aforementioned interval and then applying the given conditions, i see that the option (d) holds true.

$$ \int \frac{dy}{y}=\int (1+f^{2}(t))dt=\int (1+\frac{1}{(1+t)^{2}})dt=t-1/(1+t)+a. $$ Hence, $y(t)=ce^{t}e^{-1/(1+t)}$, where $c=e^{a}$. Now we put the value of c and see that $y$ approaches to infinity as $t$ tends to infinity

Am i correct? Is there any other better way to approach the problem?Any kind of help will be highly appreciated.Thanks everyone in advance for your time.

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I edited some "$" signs you forgot. – k1next Dec 6 '12 at 10:42
@macydanim thanks a lot. – learner Dec 6 '12 at 10:43

Sorry but one cannot choose $y$ and then adjust $f$... instead, one is given $f$ and the question is to determine the properties of the solution(s) $y$. In the present case, one may wish to show first that every solution $y$ is given by the formula $$ y(t)=y(0)\,\exp\left(\int_0^t(1+f(s)^2)\,\mathrm ds\right). $$ Once this is done, which property (properties) amongst (a)-(b)-(c)-(d) is (are) true should be clear.

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Furthermore y(t)=1/(1+t) is never a solution, for any function f. – Did Dec 6 '12 at 11:15
sorry..i have actually taken f(t) to be 1/(1+t) is just typo. – learner Dec 6 '12 at 11:19
And you solved dy/dt=(1+f(t)^2)y(t) when f(t)=1/(1+t)? How so? – Did Dec 6 '12 at 11:24
Right. This is a special case of the formula in my post, hence you should be in the position of proving it in the general case, which is necessary since, as I said, you are supposed to solve the question for every possible f. – Did Dec 6 '12 at 11:44
yes sir. But i could not find any way to prove it in the general case. – learner Dec 6 '12 at 11:50

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