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The characteristic equation

Substituting $y = e^{rx}$ turns $ay'' + by' + cy = 0$ into a quadratic, and two distinct real roots give the whole general solution.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to write the characteristic equation of a constant-coefficient second-order equation, factor it, check a factorisation against the sum and product of its roots, write the general solution when the roots are real and distinct, and fit the two constants to a pair of initial conditions.

2. What you already have

You can solve first-order equations, and you know that $e^{kx}$ is the function that reproduces itself under differentiation, multiplied by $k$. That single fact is what makes second-order constant-coefficient equations solvable by algebra rather than by integration: guess an exponential, and every derivative turns into multiplication.

3. Words this lesson uses

4. A differential equation that is really a quadratic

Try $y = e^{rx}$ in $ay'' + by' + cy = 0$. Each derivative brings down a factor of $r$, so the left-hand side becomes

$$\left(ar^{2} + br + c\right)e^{rx}.$$

An exponential is never zero, so the whole expression vanishes exactly when the bracket does. The differential equation has been replaced by a quadratic, and that replacement is the entire method.

When the quadratic has two distinct real roots $r_1 \ne r_2$, both $e^{r_1x}$ and $e^{r_2x}$ are solutions, and so is every combination

$$y = c_1e^{r_1x} + c_2e^{r_2x}.$$

That last step is linearity doing its work: the left-hand side of the equation is built from differentiation and multiplication by constants, both of which pass through a sum, so adding two solutions gives a third. The property fails immediately for a nonlinear equation, which is why classifying came first.

Two constants is exactly the right number. A second-order initial value problem prescribes $y(x_0)$ and $y'(x_0)$, which is two equations in $c_1$ and $c_2$, and because the two exponentials are independent those equations always have a solution.

Finally, the roots are worth reading rather than merely computing: they add to $-b/a$ and multiply to $c/a$. Any factorisation that fails either test is wrong, and the check costs nothing.

5. Why two constants, and not one or three

The general solution of a first-order equation carries one arbitrary constant; a second-order one carries two. The reason is not a pattern to memorise but a counting argument: to pin down a curve you need to say where it starts and how fast it is going, because the equation determines the curvature from those two and nothing else.

Concretely, with $y = c_1e^{r_1x} + c_2e^{r_2x}$ and conditions at $x = 0$:

$$c_1 + c_2 = y(0), \qquad r_1c_1 + r_2c_2 = y'(0).$$

This pair has a unique solution precisely when $r_1 \ne r_2$ — the determinant of the system is $r_2 - r_1$. So distinct roots is not a convenience; it is the condition that makes the two constants independently adjustable. When the roots coincide the system degenerates, and the next lesson repairs it.

6. What the signs of the roots tell you

Before any constants are found, the roots already describe the long-run behaviour.

And the signs are readable from the coefficients without factoring. If $a$, $b$ and $c$ are all positive, the product of the roots is positive and their sum is negative, so either both roots are negative or they are a complex pair with negative real part — in both cases everything decays. That single observation is the whole of stability for this equation, and unit 4 generalises it to systems.

7. The method, step by step, and how to check it

Every constant-coefficient equation $ay'' + by' + cy = 0$ with distinct real roots is solved by the same sequence.

1. Characteristic equation. Substitute $y = e^{rx}$. Then $y' = re^{rx}$ and $y'' = r^{2}e^{rx}$, so every term carries the factor $e^{rx}$, which is never zero and can be divided out. What is left is $ar^{2} + br + c = 0$.
2. Roots. Factor it, or use $r = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$. The discriminant $b^{2} - 4ac$ says in advance which case you are in: positive gives two real roots (this lesson), zero gives a repeated root, negative gives a complex pair (both in the next lesson).
3. General solution. $y = c_1e^{r_1x} + c_2e^{r_2x}$. Any combination works because the equation is linear and homogeneous.
4. Differentiate the general solution before using any condition: $y' = r_1c_1e^{r_1x} + r_2c_2e^{r_2x}$.
5. Conditions. Put $x = x_{0}$ into $y$ and $y'$ and set them equal to the given values. At $x_{0} = 0$ every exponential becomes $1$, leaving two linear equations.
6. Solve the pair by elimination: multiply one equation so that a constant cancels when the two are added.

How to check the roots. For $ar^{2} + br + c$ the roots always satisfy $r_1 + r_2 = -\frac{b}{a}$ and $r_1r_2 = \frac{c}{a}$. Checking both takes two lines, and it catches the common factorisation error that gets the product right and the sum wrong.

How to check the answer. Put $x = 0$ into your final $y$ and $y'$ and confirm they give the stated values. Then check one exponential in the equation: for $e^{r_1x}$ the left side is $(ar_1^{2} + br_1 + c)e^{r_1x}$, which is zero exactly because $r_1$ is a root. If the conditions fail, the usual cause is substituting $x = 0$ into $y$ before differentiating, which loses the factors $r_1$ and $r_2$ from the second equation.

8. In the world: a door closer that does not slam

A door closer is a spring that pulls the door shut and a damper filled with oil that slows it. With $\theta$ the angle the door is open, a closer is modelled by $\theta'' + c\,\theta' + k\,\theta = 0$, and its designer chooses $c$ and $k$ so that the door closes quickly without swinging past the frame and bouncing. That means two negative real roots: no oscillation, only decay.

Take $c = 5$ and $k = 6$ per second squared. The characteristic equation is $r^{2} + 5r + 6 = (r + 2)(r + 3) = 0$, so $r = -2$ and $r = -3$, and $\theta = c_1e^{-2t} + c_2e^{-3t}$. Let the door go from $90$ degrees at rest: $c_1 + c_2 = 90$ and $-2c_1 - 3c_2 = 0$, so $c_1 = 270$ and $c_2 = -180$, and

$$\theta = 270e^{-2t} - 180e^{-3t}.$$

Both terms decay; the slower one, $e^{-2t}$, sets how long closing takes. After $2$ seconds $\theta = 270e^{-4} - 180e^{-6} \approx 4.9 - 0.4 = 4.5$ degrees. With a weaker damper, $c = 2$, the roots would be complex and the door would swing through the frame and bounce, the subject of the next lesson.

9. In the world: an unstable balance, and how little time a controller has

Balance a broom upright on your palm. A small tilt $\theta$ grows, because gravity pulls a tilted broom further over: for a rod of effective length $L$, $\theta'' = \frac{g}{L}\theta$. The characteristic equation $r^{2} = \frac{g}{L}$ has roots of opposite sign, $\pm\sqrt{g/L}$, and the positive one means every tilt except exactly zero grows exponentially.

For $L = 1$ metre, $\sqrt{9.8} \approx 3.13$ per second, and a tilt grows like $e^{3.13t}$, doubling every $\frac{\ln 2}{3.13} \approx 0.22$ seconds. That number is the whole engineering specification of a self-balancing scooter or a rocket standing on its engine: the controller must sense the tilt and push back in a fraction of $0.22$ seconds, or the tilt doubles before it can act. A broom twice as long has growth rate $\sqrt{9.8/2} \approx 2.2$, doubling every $0.31$ seconds, which is why a long broom is easier to balance than a pencil.

The negative root is also there: a particular combination of tilt and tilting speed falls back to upright. A controller's job is to steer the state onto exactly that combination, the stable direction of the saddle that unit 4 draws.

10. In the world: a building on isolators during an earthquake

Buildings in earthquake zones are increasingly set on base isolators, stacks of rubber and steel that let the ground move under the building while the building moves slowly. After the shaking stops, the building's sideways offset $y$ from its base obeys $y'' + cy' + ky = 0$, and the isolators are chosen heavily damped so the offset dies away without swaying back and forth.

Suppose the isolators give $c = 1.5$ and $k = 0.5$ per second squared. Then $r^{2} + 1.5r + 0.5 = (r + 1)(r + 0.5) = 0$, so $r = -1$ and $r = -0.5$: two negative real roots, an overdamped return. The slow root, $-0.5$, governs: the offset falls by a factor of $e$ every $2$ seconds and to a twentieth of itself in about $2\ln 20 \approx 6$ seconds. Engineers read the two numbers they care about straight off the characteristic equation, before any constant is fitted: whether the building oscillates (it does not, because the roots are real) and how long it takes to settle (set by the root nearest zero). The constants only say how large the offset was when the shaking stopped.

11. The characteristic equation is not the differential equation with the primes rubbed out

Written quickly, $y'' - 5y' + 6y = 0$ becomes $r^{2} - 5r + 6 = 0$ and it looks like a notational trick — cross out the $y$s, write $r$ for each prime. It is not, and treating it as one produces two specific errors. The first is applying it to an equation with variable coefficients: $x^{2}y'' - 2y = 0$ has no characteristic equation, because substituting $e^{rx}$ leaves an $x^{2}$ behind and there is nothing to divide out. The second is forgetting that the roots are exponents, not solutions: a learner who reports "the solution is $r = 2$ and $r = 3$" has stopped one line early. The substitution is legitimate for exactly one reason — $e^{rx}$ is an eigenfunction of differentiation — and knowing that reason is what tells you when the trick is available.

12. An initial value problem, end to end

1. Solve $y'' - y' - 6y = 0$ with $y(0) = 5$, $y'(0) = 0$. Write the characteristic equation and factor it.

   $r^{2} - r - 6 = (r - 3)(r + 2) = 0 \quad\Rightarrow\quad r = 3,\ -2$

   Check: the roots add to $1$ and multiply to $-6$, as the coefficients promise.

2. Write the general solution and its derivative.

   $y = c_1e^{3x} + c_2e^{-2x}, \qquad y' = 3c_1e^{3x} - 2c_2e^{-2x}$

   Two distinct real roots give two independent exponentials.

3. Apply both initial conditions at $x = 0$.

   $$\begin{aligned} c_1 + c_2 &= 5 \\ 3c_1 - 2c_2 &= 0 \end{aligned}$$

   At $x = 0$ every exponential equals $1$: two conditions, two unknowns.

4. Solve the pair.

   $c_1 = 5 - c_2 \;\Rightarrow\; 15 - 5c_2 = 0 \;\Rightarrow\; c_2 = 3,\ c_1 = 2, \qquad y = 2e^{3x} + 3e^{-2x}$

   Substitute the first equation into the second.

5. Read the long-run behaviour.

   $x \to \infty: \quad 2e^{3x} \to \infty, \quad 3e^{-2x} \to 0$

   The solution starts with zero slope and still grows without limit: the positive root wins.

13. Reading a factorisation for errors

1. Someone factors $r^{2} + 7r + 12$ as $(r + 2)(r + 6)$, giving roots $-2$ and $-6$. Check the product of the roots.

   $(-2)(-6) = 12 = \dfrac{c}{a} \quad \checkmark$

   The product matches, which is why the mistake looks plausible.

2. Check the sum of the roots as well.

   $(-2) + (-6) = -8 \ne -7 = -\dfrac{b}{a}$

   The roots must add to minus the middle coefficient, and these do not, so the factorisation is wrong.

3. Find the pair that satisfies both conditions.

   $(-3) + (-4) = -7, \quad (-3)(-4) = 12 \quad\Rightarrow\quad (r + 3)(r + 4)$

   List the factor pairs of $12$ and pick the one that adds to $7$.

4. Confirm by expanding the new factorisation.

   $(r + 3)(r + 4) = r^{2} + 4r + 3r + 12 = r^{2} + 7r + 12 \quad \checkmark$

   Expanding is the definitive check: it must reproduce the polynomial exactly.

5. Write the general solution and read its behaviour.

   $y = c_1e^{-3x} + c_2e^{-4x} \;\to\; 0 \text{ as } x \to \infty$

   Both roots are negative, so every solution decays. The wrong factorisation would have given the same verdict and the wrong rates, which is why checking both conditions matters.

14. Roots from the quadratic formula

1. Solve $y'' + 2y' - 2y = 0$ with $y(0) = 0$, $y'(0) = 2\sqrt{3}$. The characteristic equation does not factor over the integers, so use the formula.

   $r = \dfrac{-2 \pm \sqrt{2^{2} - 4 \cdot 1 \cdot (-2)}}{2} = \dfrac{-2 \pm \sqrt{12}}{2}$

   $r = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$ with $a = 1$, $b = 2$, $c = -2$. The discriminant is $12 > 0$, so the roots are real and distinct.

2. Simplify the surd and divide through by $2$.

   $\sqrt{12} = 2\sqrt{3} \quad\Rightarrow\quad r = -1 \pm \sqrt{3}$

   $12 = 4 \times 3$ and $\sqrt{4} = 2$.

3. Write the general solution and its derivative.

   $y = c_1e^{(-1 + \sqrt{3})x} + c_2e^{(-1 - \sqrt{3})x}, \qquad y' = (-1 + \sqrt{3})c_1e^{(-1 + \sqrt{3})x} + (-1 - \sqrt{3})c_2e^{(-1 - \sqrt{3})x}$

   Irrational roots change nothing about the method.

4. Put $x = 0$ into $y$ and use $y(0) = 0$.

   $c_1 + c_2 = 0 \quad\Rightarrow\quad c_2 = -c_1$

   Every exponential equals $1$ at $x = 0$.

5. Put $x = 0$ into $y'$, use $y'(0) = 2\sqrt{3}$, and substitute $c_2 = -c_1$.

   $(-1 + \sqrt{3})c_1 - (-1 - \sqrt{3})c_1 = 2\sqrt{3} \quad\Rightarrow\quad 2\sqrt{3}\,c_1 = 2\sqrt{3}$

   The $-1$ terms cancel and the $\sqrt{3}$ terms add.

6. Divide both sides by $2\sqrt{3}$.

   $c_1 = 1, \qquad c_2 = -1, \qquad y = e^{(-1 + \sqrt{3})x} - e^{(-1 - \sqrt{3})x}$

   The difference of the roots, $2\sqrt{3}$, is exactly what the two conditions divide by, as it always is.

7. Read the long-run behaviour from the signs of the roots.

   $-1 + \sqrt{3} \approx 0.73 > 0, \qquad -1 - \sqrt{3} \approx -2.73 < 0$

   Opposite signs: the second term dies and the first grows, so the solution grows like $e^{0.73x}$. The product of the roots, $-2$, is negative, which predicted opposite signs before any root was found.

15. Your turn: solve $2y'' - 5y' + 2y = 0$ with $y(0) = 3$, $y'(0) = 0$

1. Write and factor the characteristic equation.

   $2r^{2} - 5r + 2 = (2r - 1)(r - 2) = 0 \quad\Rightarrow\quad r = \tfrac{1}{2},\ 2$

   Check: they add to $\tfrac{5}{2}$ and multiply to $1$, matching $-b/a$ and $c/a$.

2. Your turn: work this step out. Its working is at the end of the packet.

   Write the solution, differentiate, and apply the conditions.

3. Your turn: work this step out. Its working is at the end of the packet.

   Solve for the constants.

16. Guided practice

Each of these equations has two whole-number exponents in its general solution. Match each one to its pair of roots.

17. Guided practice

Complete the worked solution of $y'' + y' - 6y = 0$ with $y(0) = 35$ and $y'(0) = 0$.

1. Write the characteristic equation and factor it.

   $r^{2} + r - 6 = (r - 2)(r + 3) = 0 \quad\Rightarrow\quad r = 2,\ -3$

   Two numbers adding to $-1$ and multiplying to $-6$.

2. Write the general solution and its derivative.

   $y = c_1e^{2x} + c_2e^{-3x}, \qquad y' = 2c_1e^{2x} - 3c_2e^{-3x}$

   Two distinct real roots give two exponentials.

3. Substitute $x = 0$ into both.

   $$\begin{aligned} c_1 + c_2 &= 35 \\ 2c_1 - 3c_2 &= 0 \end{aligned}$$

   At $x = 0$ every exponential equals $1$.

4. From the second equation, $c_2 = \frac{2}{3}c_1$; substitute into the first and solve for $c_1$.

   $c_1\left(1 + \dfrac{2}{3}\right) = 35 \quad\Rightarrow\quad c_1 =$ _a

   Multiply both sides by $\frac{3}{5}$.

5. Substitute back for $c_2$.

   $c_2 = \dfrac{2}{3} \times c_1 =$ _b

   Use the relation from the second equation.

18. Guided practice

Which of these functions is a solution of $y'' + 3y' + 2y = 0$?

• $y = e^{2x}$
• $y = e^{-2x}$
• $y = e^{3x}$
• $y = e^{-3x}$

19. Practice

Substitute $y = e^{rx}$ into $3y'' + 5y' + 7y = 0$ and divide by $e^{rx}$. Write the polynomial in $r$ that is left.

Answer:

20. Practice

For $y'' - 7y' + 10y = 0$, what is the larger root of the characteristic equation?

Answer:

21. Practice

Before its motors react, the tilt $y$ of a self-balancing robot, $x$ seconds after a knock, obeys $y'' - 2y' - 8y = 0$: one mode of the tilt dies away and one grows. Solve $y'' - 2y' - 8y = 0$ with $y(0) = 5$ and $y'(0) = 8$. Write $y$ as a formula in $x$ (type each exponential as e^(...)).

Answer:

22. Somewhere new

Without solving any of them, fill in the sum of the roots, their product and the discriminant for each equation.

23. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

24. Test question

Before its motors react, the tilt $y$ of a self-balancing robot, $x$ seconds after a knock, obeys $y'' - 2y' - 15y = 0$: one mode of the tilt dies away and one grows. Solve $y'' - 2y' - 15y = 0$ with $y(0) = 3$ and $y'(0) = 7$. Write $y$ as a formula in $x$ (type each exponential as e^(...)).

Answer:

25. What you can do now

You can turn a constant-coefficient equation into its characteristic quadratic, factor it, and build the general solution from two distinct real roots. Say in your own words why the substitution $y = e^{rx}$ is legitimate at all.

Working for the steps left to you

15. Your turn: solve $2y'' - 5y' + 2y = 0$ with $y(0) = 3$, $y'(0) = 0$, step 2

15. Your turn: solve $2y'' - 5y' + 2y = 0$ with $y(0) = 3$, $y'(0) = 0$, step 3