Second-Order Linear Homogeneous ODEs

≈ 30 min

A second-order linear homogeneous ODE with constant coefficients has the form

a y'' + b y' + c y = 0.

Trying $y = e^{rx}$ reduces it to the characteristic equation

a r^2 + b r + c = 0.

The roots decide the solution:

Distinct real roots $r_1, r_2$ : $\;y = A e^{r_1 x} + B e^{r_2 x}.$
Repeated root $r$ : $\;y = (A + Bx) e^{r x}.$
Complex roots $\alpha \pm \beta i$ : $\;y = e^{\alpha x}\big(A\cos\beta x + B\sin\beta x\big).$

Worked example. $y'' - 5y' + 6y = 0$ gives $r^2 - 5r + 6 = (r-2)(r-3) = 0$ , so $y = A e^{2x} + B e^{3x}.$

What is the characteristic equation of $y'' - 5y' + 6y = 0$ ?

$r^2 - 5r + 6 = 0$

$r^2 + 5r + 6 = 0$

$6r^2 - 5r + 1 = 0$

$r^2 - 5r - 6 = 0$

The characteristic equation of $y'' - 5y' + 6y = 0$ factors as $(r-2)(r-3)=0$ . What is the general solution?

$y = A e^{2x} + B e^{3x}$

$y = A e^{-2x} + B e^{-3x}$

$y = (A + Bx) e^{2x}$

$y = e^{2x}(A\cos 3x + B\sin 3x)$

What is the general solution of $y'' + 4y = 0$ ?

$y = A\cos 2x + B\sin 2x$

$y = A e^{2x} + B e^{-2x}$

$y = (A + Bx)e^{2x}$

$y = A e^{4x}$

The equation $y'' - 6y' + 9y = 0$ has a repeated root. Find that root $r$ .