I want to make sure I'm solving these differential equations correct. Is this right?
- $$y" - 2y' - 3y = 0$$ with initial conditions: $y(0) = 2$ and $y'(0) = 2$
so the auxiliary equation:
$$r^2 - 2r -3 = 0$$ $$(r + 1)(r-3) = 0$$ so $r = -1, 3$
so generally: $$y = c_1e^{-x} + c_2e^{3x}$$
if $y(0) = 2 = c_1 + c_2$ and $y'(0) = 2 = -c_1 + 3c_2$ then $c_2 = 1$ and $c_1 = 1$
so the solution given the initial conditions is:
$$y = e^{-x} + e^{3x}$$
Is that right?
- I'm a bit stuck on this one:
$$3y" + 4y' - 3y = 0$$
the auxiliary equation is:
$$3r^2 + 4r -3 = 0$$
The roots can't be found by factoring easily so I'll use the quadratic:
$$r = \frac{-4 \pm \sqrt{16 - 4(3)(-3)}}{6}$$
$$r = \frac{-4 \pm \sqrt{16 + 36}}{6}$$
$$r = \frac{-2}{3} \pm \frac{2 \sqrt{13}}{3}$$
so is the solution just:
$$y = c_1e^{r_1x} + c_2e^{r_2x}$$
where r1 and r2 are just the two roots found by solving the quadratic?