<s>[INST] Solving second-order differential equations. I have a few problems that I'm trying to work through. Want to see if these few are correct.


*

*$$3y'' + 4y' - 3y = 0$$
auxiliary equation is: $$3r^2 + 4r -3 = 0$$ where $a = 3$, $b = 4$, $c = -3$
can't really find roots by factoring so gonna use quadratic:
$$r = \frac{-4 \pm \sqrt{16 - 4(3)(-3)}}{6}$$
$$r = \frac{-4 \pm \sqrt{52}}{6}$$
$$r = \frac{-2}{3} \pm \frac{\sqrt{52}}{6}$$
$$r = \frac{-2}{3} \pm \frac{2\sqrt{13}}{3}$$
so there are 2 real roots. So the general solution is:
$$y = c_1e^{r_1x} + c_2e^{r_2x}$$
where $r_1 = \frac{-2}{3} + \frac{2\sqrt{13}}{3}$
where $r_2 = \frac{-2}{3} - \frac{2\sqrt{13}}{3}$


*$$9y'' + 4y = 0$$
auxiliary equation (could have used quadratic): $$9r^2 + 4 = 0$$
$$9r^2 = -4$$
$$r^2 = -4/9$$
$$r = \pm \frac{2}{3}i$$
so the two roots are:
$$r_1 = 0 + \frac{2}{3}i$$ and $$r = 0 - \frac{2}{3}i$$
where $\alpha = 0$ and $\beta = \frac{2}{3}$
and so the general solution is:
$$y = e^0(c_1cos\frac{2}{3}x + c_2sin\frac{2}{3}x)$$


*$$y = y''$$
$$y'' - y = 0$$
$$r^2 - 1 = 0$$
$$r^2 = 1$$
$$r = \pm 1$$
two real roots so:
general solution is: $$y = c_1e^x + c_2e^{-x}$$


*$$y'' + 2y = 0$$
$$r^2 + 2 = 0$$
$$r^2 = -2$$
$$r = \pm \sqrt{-2} = \pm \sqrt{2}i$$ and so $\alpha = 0$ and $\beta = \sqrt{2}$ and so $e^{\alpha x} = 1$
so
$$y = c_1cos\sqrt{2}x + c_2sin\sqrt{2}x$$
do these look right?
 [/INST] For this type of problems you can also follow the table given below.

I think, this table will help you to deal with all linear differential equations with constant coefficients. 
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