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The general solution to the equation $y''+by'+cy=0$ approaches to $0$ as $x$ approaches infinity if

  • $b$ is negative $c$ is positive
  • $b$ is positive $c$ is negative
  • $b$ is positive $c$ is positive
  • $b$ is negative $c$ is negative
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    The first bullet has an $a$ in it. Is this supposed to be a question? What is it and where are you stuck? – 2017-01-03
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    @B.Goddard maybe the reason for $a=1$ is that the second order term does not vanish. – 2017-01-03
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    @Cuteboy There's no $a$ in the equation. – 2017-01-03
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    @B.Goddard no, but I mean that one can always rescale the equation if $a\neq 0$. – 2017-01-03
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    Perhaps you are reciting a multiple choice question here, but this is not the sort of content that Math.SE seeks to accumulate. Please review [ask] and state a question in your own words. – 2017-01-03
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    @Cuteboy I'm not talking about the equation, but about the first bullet point, which, before the edit, had an $a$ in it. He has fixed it now. – 2017-01-03
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    @B.Goddard ok, i also saw that^^, just automatically corrected in mind – 2017-01-03

2 Answers 2

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Hint: to find the solutions of this differential equation, replace $y''$ by $r^2$ and $y'$ by $r$, and solve the quadratic polynomial $$r^2 + br + c = 0$$ to find the roots $r_1,r_2$.

The solution $y(x)$ of the differential equation is then given by $$y(x) = Ae^{r_1x} + Be^{r_2x}$$ if $r_1 \neq r_2$, and otherwise by $$y(x) = Ae^{r_1x} + Bxe^{r_1x}.$$ Note that if $r_1,r_2$ are complex, they are necessarily complex conjugates of the form $\alpha \pm \beta i$, and you can rewrite the solution as $$y(x) = e^{\alpha x}(A\cos(\beta x) + B \sin (\beta x)).$$

The signs of $b$ and $c$ will determine the signs of $r_1$ and $r_2$, which in turn determine the behaviour of your function as $x \to \infty$.

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    I dont really think the signs of $b$ and $c$ really matter, but indeed the term $b^2-4c$. – 2017-01-03
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    @Cuteboy The discriminant determines which of the three solutions you get, but the signs of $r_1$ and $r_2$ are determined by $-b \pm \sqrt{b^2 - 4c}$. And, if the signs didn't matter, the question itself wouldn't make much sense, would it? – 2017-01-03
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    you are right, but for me the question just does not look appropriate, since in fact the roots play an essential role but the sign itself. – 2017-01-03
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    @Cuteboy Let's assume $\sqrt{b^2-4c} > 0$. If $c > 0$ then $\sqrt{b^2-4c} < |b|$. Then the sign of $b$ alone determines the signs of $r_1$ and $r_2$. If on the other hand $c < 0$, then $\sqrt{b^2-4c} > |b|$ and $r_1$ and $r_2$ will have a different sign. You can do this for other cases too, but in the end the signs of $b$ and $c$ will be decisive. – 2017-01-03
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    Yes, I totally agree, but I mean that just looks uncomfortable for me since if one knows some fundamentals in ODE, one will always concern the roots themself but not really the signs of the constants... – 2017-01-03
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To seek a non-zero solution, put $y=e^{mx}$ ($m$ is arbitrary constant) in the given ODE

Then the auxiliary equation you get is $e^{mx}(m^2+bm+c)=0$

or $m^2+bm+c=0$ as $e^{mx}\neq 0$.

In case of real roots,what you want is that this quadratic equation must not have a positive real root for which $b>0$ and $c>0$ (by Descarte's rule of signs) is sufficient.

In case no real root exists, even then you need the real part of the root to be negative for which you must have $b>0$.

In both cases, option $3$ is the best choice.