Linear second order equations with constant coefficients

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Let us have a look at the equations of the form

$y'' + ay' + by = 0$

(This, by the way, is also called a homogenous equation.) Except for the first term, it has the same form as

$ay' + by = 0$

We can rewrite that as

$\frac{{dy}}{y} = - \frac{b}{a}dx$

Integrating this gives us

$\ln |y| = - \frac{b}{a}x + {C_1}$

$|y| = {e^{ - \frac{b}{a}x + {C_1}}} = {C_2}{e^{ - \frac{b}{a}x}}$

The sign can be absorbed into the constant, finally giving us

$y = C{e^{ - \frac{b}{a}x}}$

So it may be an idea to test if the same type of solution could be solving our initial equations. Let us assume we have a solution of the form

$y = C{e^{rx}}$

We then have

$y' = Cr{e^{rx}},\quad y'' = C{r^2}{e^{rx}}$

Substituting this into our initial equation gives us

$C{r^2}{e^{rx}} + aCr{e^{rx}} + bC{e^{rx}} = 0$

This could be factorized as

$C{e^{rx}}\left( {{r^2} + ar + b} \right) = 0$

So if r has a value such that the second factor is 0 then we have a solution. The equation

${r^2} + ar + b = 0$

is called the characteristic equation, and the solutions of our original equation will depend on the solutions to the characteristic equation.

An example

Say we have

$y'' + 2y' - 15y = 0$

That gives the characteristic equation

${r^2} + 2r - 15 = 0$

We can solve that by (for example) splitting -15 into two factors that add up to 2. Doing so we get

$(r - 3){(r + 5)^2} = 0$

with the solutions r=3 and r=-5. We will thus have as solutions to our differential equation

$y = C{e^{3x}},\quad y = D{e^{ - 5x}}$

where C and D are constants. Matter a fact, as you can quite easily verify, we have

$y = C{e^{3x}} + D{e^{ - 5x}}$

as a solution (as matter a of fact it is the general solution). But a quadratic equation, in this case the characteristic equation) may have two real roots, two equal real roots or two complex roots. So what do these different cases correspond to? Let us have a look at the following few pages.

Up a level : Differential Equations
Previous page : Graphing all solutions
Next page : LSO -Two different real roots

Last modified: Dec 28, 2023 @ 17:32