Solution to an ordinary differential equation 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

 A: 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 $x \to \infty$. 
A: 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.
