I am of the conviction that the only true option for this question is option 1 . the solution is of the form y-3= (y-1)e^(2x/c). Clearly as x tends to infinity y will also tend to infinity and hence the solution is not bounded above. Correct me if I am wrong.
query on solution of a differential equation
1
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ordinary-differential-equations
2 Answers
1
It is important to express $y$ only in terms of $x$ while checking the limit of $y$ as $x \to \infty $.
We can simplify our expression to: $$\frac {y-3}{y-1} = e^{\frac {2x}{c}}$$ $$\Rightarrow 1- e^{\frac {2x}{c}} = \frac {2}{y-1} $$ $$\Rightarrow y = \frac {2}{1-e^{\frac {2x}{c}}}+1$$
Now as $x \to \infty $, $y \to ? $. Hope you can take it from here.
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0If x tends to + infinity then its yes 1 and if x tends to - infinity its 3 – 2017-02-18
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0@shadowkh Yes, you are right. – 2017-02-18
0
You could also just check the sign of the derivative. Then you get growth on $y\in[3,\infty)$ and $y\in(-\infty,1]$ while the solution falls for $y\in[1,3]$. As solutions are unique, the constant solutions $y\equiv 1$ and $y\equiv 3$ are boundaries for any solution starting in $(1,3)$, and as the solution falls it must tend to $1$.