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.

1. $$3y'' + 4y' - 3y = 0$$

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 $$r_2 = \frac{-2}{3} - \frac{2\sqrt{13}}{3}$$

1. $$9y'' + 4y = 0$$

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:

$$y = e^0(c_1cos\frac{2}{3}x + c_2sin\frac{2}{3}x)$$

1. $$y = y''$$ $$y'' - y = 0$$

$$r^2 - 1 = 0$$ $$r^2 = 1$$ $$r = \pm 1$$

two real roots so:

general solution is: $$y = c_1e^x + c_2e^{-x}$$

1. $$y'' + 2y = 0$$ $$r^2 + 2 = 0$$ $$r^2 = -2$$ $$r = \pm \sqrt{-2} = \pm \sqrt{2}i$$ 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?

• Yes, everything looks good. – nmasanta May 4 at 3:33
• The general solution in Example-3 can also be written as $y = c_{1} \cosh x + c_{2} \sinh x$ – nmasanta May 4 at 3:42 • If there's 1 real root, don't we need to also multiply by $e^{rx}$ by x to get the second solution? – Jwan622 May 4 at 14:00
• Yes, for one real root you have to follow $1(i)$, as there is no other root. For real equal root (say $m_{1} = m_{2} = m$), you have to follow $2(i)$. Suppose you have a differential equation $y'' - 2y' + y = 0$, then you have two roots of auxiliary equation which are equal (i.e., $m=1$), then your solution be $y = a e^{x} + b x e^{x}$ i.e., $y = (a + b x) e^{x}$. Again suppose you have a differential equation $y' - y = 0$, then you have only one root of auxiliary equation which is $m=1$, then your solution be $y = a e^{x}$. – nmasanta May 4 at 14:24