Two distinct real roots · Differential Equations

Intuition

When the characteristic equation has two different real roots, the equation has two exponential solutions, one growing or decaying at each rate, and every solution is a combination of them. The long run is read off the signs: two negative roots mean every solution dies away, a positive root means almost every solution grows, and the larger root wins in the end.

Two savings accounts, one at a higher rate. Whatever the starting amounts, after long enough the higher-rate account holds almost everything, and the total grows at its rate.

Solutions of $y''+3y'+2y=0$ , whose roots are $-1$ and $-2$ . Every solution is $C_{1}e^{-x}+C_{2}e^{-2x}$ , so every one dies away, and it can cross zero at most once on the way.

Two exponentials, and which one wins

If $ar^{2}+br+c=0$ has real roots $r_{1}\ne r_{2}$ , the general solution of is . The two exponentials are linearly independent, since their ratio is not constant, and the lesson on the Wronskian shows that two independent solutions give every solution.

y=C_{1}e^{r_{1}x}+C_{2}e^{r_{2}x}

Reading the solutions

$y''-y'-6y=0$ : roots $3$ and , so .

The larger rate dominates

r_{1}>r_{2},\ C_{1}\ne0\ \Longrightarrow\ \frac{y}{C_{1}e^{r_{1}x}}\to1

Take the exponential with the larger rate out as a factor. What remains in the bracket is its coefficient plus the other coefficient times an exponential with a negative rate, and that exponential tends to zero. So the solution divided by its leading term tends to one: in the long run the solution looks like its term with the larger rate, provided that term is present.

Proof steps

y=e^{r_{1}x}\bigl(C_{1}+C_{2}e^{(r_{2}-r_{1})x}\bigr)

Take the exponential with the larger rate out as a factor.

r_{2}-r_{1}<0\ \Longrightarrow\ e^{(r_{2}-r_{1})x}\to0

The remaining exponent is negative, so its exponential tends to zero.

C_{1}+C_{2}e^{(r_{2}-r_{1})x}\to C_{1}

So the bracket tends to the coefficient of the leading term.

\frac{y}{C_{1}e^{r_{1}x}}\to1

Dividing by the leading term, the solution behaves like it in the long run.

Applications

Practice

Two Roots, Two Exponentials

Each real root gives an exponential solution, and the general solution combines them.

r_{1}\ne r_{2}\ \Longrightarrow\ y=C_{1}e^{r_{1}x}+C_{2}e^{r_{2}x}

Try it

What is the general solution of $y''-y'-6y=0$ ?