Chapter 19

Hard

Differential Equations — Second Order and Systems

00 · Symbol Glossary

$ay''+by'+cy=0$second-order linear homogeneous equation with constant coefficients

An equation involving $y$ , $y'$ , and $y''$ , with $a,b,c$ fixed real constants and $a\neq0$ . "Homogeneous" means every term contains $y$ or a derivative of $y$ — no leftover term depending on $x$ alone.

$ar^2+br+c=0$characteristic equation

An ordinary quadratic equation in a new unknown $r$ , built directly from the coefficients of the differential equation. Its roots determine the shape of every solution — introduced in Section 01.

$r_1, r_2$roots of the characteristic equation

The two solutions of the characteristic equation (found by factoring or the quadratic formula, exactly as in ordinary algebra). Each root $r$ produces a building-block solution $e^{rx}$ .

$\Delta = b^2-4ac$discriminant

The same quantity from the quadratic formula that determines whether $r_1,r_2$ are real and distinct, real and repeated, or complex — Section 02 covers all three cases.

$c_1, c_2$the two free constants of a second-order general solution

Second-order equations require two arbitrary constants in the general solution (matching two rounds of integration, one for each order), unlike the single $C$ of Chapter 18's first-order equations.

01 · Second-Order Linear Equations and the Characteristic Equation

Chapter 18 solved equations built from $y$ and $y'$ . Adding $y''$ requires a new idea: guessing a specific exponential shape for the solution, then finding out exactly which exponential works.

Definition — Characteristic Equation

For $ay''+by'+cy=0$ , try a solution of the form $y=e^{rx}$ . Then $y'=re^{rx}$ and $y''=r^2e^{rx}$ (Chapter 03's chain rule, applied twice). Substituting:

ar^2e^{rx}+bre^{rx}+ce^{rx}=0 \implies e^{rx}(ar^2+br+c)=0

Since $e^{rx}$ is never $0$ , this requires $ar^2+br+c=0$ — the characteristic equation, an ordinary quadratic in $r$ .

Step-by-step — Solve $y''-5y'+6y=0$

Write the characteristic equation: matching coefficients

a=1,b=-5,c=6

r^2-5r+6=0

Factor:

r^2-5r+6=(r-2)(r-3)

, since

(-2)(-3)=6

and

-2+(-3)=-5

match the coefficients.

Solve for the roots:

r=2

r=3

Build one solution per root:

y_1=e^{2x}

y_2=e^{3x}

Assemble the general solution using two free constants:

y = c_1e^{2x}+c_2e^{3x}

Verify:

y'=2c_1e^{2x}+3c_2e^{3x}

;

y''=4c_1e^{2x}+9c_2e^{3x}

. Check

y''-5y'+6y = (4c_1e^{2x}+9c_2e^{3x}) - 5(2c_1e^{2x}+3c_2e^{3x}) + 6(c_1e^{2x}+c_2e^{3x})

. Collecting the

e^{2x}

terms:

4c_1-10c_1+6c_1=0

✓. Collecting the

e^{3x}

terms:

9c_2-15c_2+6c_2=0

✓.

02 · Repeated and Complex Roots

The characteristic equation is an ordinary quadratic, so its discriminant $\Delta=b^2-4ac$ determines which of three cases applies.

Definition — Three Cases by Discriminant

$\Delta>0$ : two distinct real roots $r_1,r_2$ . General solution: $y=c_1e^{r_1x}+c_2e^{r_2x}$ (Section 01).
$\Delta=0$ : one repeated root $r$ . General solution: $y = c_1e^{rx}+c_2xe^{rx}$ .
$\Delta<0$ : complex roots $r=\alpha\pm\beta i$ . General solution: $y=e^{\alpha x}\big(c_1\cos(\beta x)+c_2\sin(\beta x)\big)$ .

Step-by-step — Solve the repeated-root case $y''-6y'+9y=0$

Characteristic equation:

r^2-6r+9=0

Compute the discriminant:

\Delta = (-6)^2-4(1)(9) = 36-36=0

— repeated root case.

Solve (factor as a perfect square):

r^2-6r+9=(r-3)^2=0 \implies r=3

(a double root).

Apply the repeated-root formula from Section 02:

y = c_1e^{3x}+c_2xe^{3x}

❌ Writing c1·e^(rx) + c2·e^(rx) for a repeated root

On the same equation, a tempting but incorrect shortcut: since there's only one root $r=3$ , writing $y=c_1e^{3x}+c_2e^{3x}$ .

Why it breaks: $c_1e^{3x}+c_2e^{3x} = (c_1+c_2)e^{3x}$ — algebraically, this is just one arbitrary constant ( $c_1+c_2$ combined) multiplying $e^{3x}$ , not two genuinely independent constants. A second-order equation's general solution needs two constants that cannot be merged into one, matching the two rounds of integration a second-order equation represents (Section 00's symbol glossary note).

Consequence: with only one effective constant, this incorrect form cannot satisfy an arbitrary pair of initial conditions $y(0)=y_0$ , $y'(0)=y_0'$ — plugging in generally produces a contradiction unless $y_0'$ happens to equal $3y_0$ . The correct extra building-block solution is $xe^{3x}$ , not a second copy of $e^{3x}$ ; verify $\dfrac{d}{dx}[xe^{3x}] = e^{3x}+3xe^{3x}$ (product rule, Chapter 02) actually satisfies the original equation, which a repeated $e^{3x}$ term cannot independently contribute.

Step-by-step — Solve the complex-root case $y''+4y=0$

Characteristic equation:

r^2+0r+4=0

, i.e.

r^2=-4

Compute the discriminant:

\Delta = 0^2-4(1)(4)=-16<0

— complex root case.

Solve for $r$ :

r=\pm\sqrt{-4} = \pm2i

, so

\alpha=0

\beta=2

(matching

r=\alpha\pm\beta i

Apply the complex-root formula from Section 02:

y = e^{0\cdot x}\big(c_1\cos(2x)+c_2\sin(2x)\big) = c_1\cos(2x)+c_2\sin(2x)

Verify using Chapter 02, Section 09's trig derivatives:

y'=-2c_1\sin(2x)+2c_2\cos(2x)

(chain rule, Chapter 03);

y''=-4c_1\cos(2x)-4c_2\sin(2x)

. Check

y''+4y = (-4c_1\cos2x-4c_2\sin2x)+4(c_1\cos2x+c_2\sin2x) = 0

✓.

Why sine and cosine appear from complex roots

Substituting $r=\beta i$ directly gives the building-block solution $e^{i\beta x}$ , a complex-valued expression — not itself one of the two real solutions wanted here. The identity $e^{i\beta x} = \cos(\beta x)+i\sin(\beta x)$ (stated here as a given fact, not derived) shows that the real and imaginary parts of $e^{i\beta x}$ , namely $\cos(\beta x)$ and $\sin(\beta x)$ , are themselves both real-valued solutions of the original equation — which is why they appear directly in the general solution instead of the complex exponential itself.

03 · Nonhomogeneous Equations — A Compressed Look

Definition — Nonhomogeneous Equation and Particular Solution

$ay''+by'+cy=g(x)$ with $g(x)\neq0$ is nonhomogeneous. Its general solution is $y=y_h+y_p$ , where $y_h$ solves the corresponding homogeneous equation ( $g(x)=0$ , Sections 01–02) and $y_p$ is any single solution of the full equation, typically found by guessing a form matching $g(x)$ and solving for its coefficients.

Example — A single worked case, kept compressed

For $y''-5y'+6y=12$ (same left side as Section 01's example), guess a constant particular solution $y_p=A$ (since the right side is a constant): $y_p'=0$ , $y_p''=0$ , so the equation becomes $0-0+6A=12 \implies A=2$ . Combined with Section 01's homogeneous solution: $y = c_1e^{2x}+c_2e^{3x}+2$ .

04 · Systems of First-Order Equations — Solving by Elimination

Some problems involve two unknown functions changing together, each one's rate of change depending on both. This is solved here using substitution and elimination — reducing the system to a single, higher-order equation in one unknown, exactly the kind already solved in Sections 01–02 — rather than any specialized machinery for handling several equations simultaneously.

Definition — System of First-Order Equations and the Elimination Method

A system $\dfrac{dx}{dt} = ax+by$ , $\dfrac{dy}{dt}=cx+dy$ (constants $a,b,c,d$ ) can be reduced to one equation in $x$ alone: differentiate the first equation with respect to $t$ , then substitute the second equation (and the first, solved for $y$ ) to eliminate $y$ and $y'$ entirely.

Step-by-step — Solve $\dfrac{dx}{dt}=x+2y,\ \dfrac{dy}{dt}=2x+y$ by elimination

Differentiate the first equation with respect to $t$ :

x'' = x'+2y'

Substitute the second equation ( $y'=2x+y$ ) to remove $y'$ :

x'' = x'+2(2x+y) = x'+4x+2y

Solve the first original equation for $y$ in terms of $x,x'$ : from

x'=x+2y

y = \dfrac{x'-x}{2}

Substitute this expression for $y$ into step 2's equation, removing $y$ entirely:

x'' = x'+4x+2\left(\dfrac{x'-x}{2}\right) = x'+4x+x'-x = 2x'+3x

Rearrange into standard second-order form:

x''-2x'-3x=0

Solve using Section 01's characteristic-equation method:

r^2-2r-3=0 \implies (r-3)(r+1)=0 \implies r=3

r=-1

. So

x(t) = c_1e^{3t}+c_2e^{-t}

Recover $y(t)$ using step 3's relation $y=\dfrac{x'-x}{2}$ :

x'(t) = 3c_1e^{3t}-c_2e^{-t}

. So

y = \dfrac{(3c_1e^{3t}-c_2e^{-t})-(c_1e^{3t}+c_2e^{-t})}{2} = \dfrac{2c_1e^{3t}-2c_2e^{-t}}{2} = c_1e^{3t}-c_2e^{-t}

Verify against the second original equation, $y'=2x+y$ :

y' = 3c_1e^{3t}+c_2e^{-t}

(differentiating step 7's result). Compute

2x+y = 2(c_1e^{3t}+c_2e^{-t}) + (c_1e^{3t}-c_2e^{-t}) = 3c_1e^{3t}+c_2e^{-t}

— matches

y'

exactly. ✓

❌ Stopping the elimination before every trace of the second unknown is removed

On the same system, a common error at step 4: substituting the second equation for $y'$ (step 2) but forgetting to also replace the leftover bare $y$ term, leaving an equation like $x''=x'+4x+2y$ still containing $y$ .

Why it breaks: the whole point of elimination is to end with a single equation in a single unknown function — an equation still containing both $x$ and $y$ cannot be solved by Section 01's characteristic-equation method, which only applies to one function of one variable.

Consequence: step 3's extra substitution (solving the original first equation for $y$ itself, not just $y'$ ) is not optional busywork — it's the step that finishes removing every trace of $y$ . Skipping it leaves a mixed equation that looks superficially similar to a solvable one but cannot actually be solved by the tools in this chapter.

05 · Exercises

EXERCISE 19.1

Form the characteristic equation from the coefficients, factor it, and combine the two exponential building-block solutions.

$y''-y'-6y=0$ : $r^2-r-6=0 \implies (r-3)(r+2)=0 \implies r=3,-2$ . General solution: $y=c_1e^{3x}+c_2e^{-2x}$ .

Solve $y''-y'-6y=0$ .

EXERCISE 19.2

Compute the discriminant first to identify which of the three cases applies, then build the general solution using the matching formula from Section 02.

$y''+4y'+4y=0$ : $r^2+4r+4=0$ , $\Delta=16-16=0$ — repeated root. $(r+2)^2=0 \implies r=-2$ . General solution: $y=c_1e^{-2x}+c_2xe^{-2x}$ .

Solve $y''+4y'+4y=0$ .

EXERCISE 19.3

Find the complex roots via the quadratic formula, identify $\alpha$ and $\beta$ , then apply the complex-root general solution formula.

$y''-2y'+5y=0$ : $r^2-2r+5=0$ . Quadratic formula: $r = \dfrac{2\pm\sqrt{4-20}}{2} = \dfrac{2\pm\sqrt{-16}}{2} = 1\pm2i$ . So $\alpha=1$ , $\beta=2$ . General solution: $y = e^x(c_1\cos2x+c_2\sin2x)$ .

Solve $y''-2y'+5y=0$ .

EXERCISE 19.4

Use the initial conditions to solve for both constants — you'll need both $y(0)$ and $y'(0)$ to pin down two constants.

From Exercise 19.1, $y=c_1e^{3x}+c_2e^{-2x}$ , $y'=3c_1e^{3x}-2c_2e^{-2x}$ . Given $y(0)=1$ , $y'(0)=8$ : $c_1+c_2=1$ and $3c_1-2c_2=8$ . From the first, $c_2=1-c_1$ . Substitute: $3c_1-2(1-c_1)=8 \implies 3c_1-2+2c_1=8 \implies 5c_1=10 \implies c_1=2$ , so $c_2=-1$ . Particular solution: $y=2e^{3x}-e^{-2x}$ .

Solve the initial value problem $y''-y'-6y=0$ , $y(0)=1$ , $y'(0)=8$ (building on Exercise 19.1).

EXERCISE 19.5

Differentiate one equation, substitute the other to remove its derivative, then solve the remaining equation for the other unknown to remove every trace of it too.

$\dfrac{dx}{dt}=4x-2y$ , $\dfrac{dy}{dt}=x+y$ . Differentiate the first: $x''=4x'-2y'$ . Substitute $y'=x+y$ : $x''=4x'-2(x+y)=4x'-2x-2y$ . Solve the first equation for $y$ : $2y=4x-x' \implies y=2x-\dfrac{x'}{2}$ . Substitute: $x''=4x'-2x-2\left(2x-\dfrac{x'}{2}\right)=4x'-2x-4x+x'=5x'-6x$ . Rearrange: $x''-5x'+6x=0$ . Characteristic equation: $r^2-5r+6=0 \implies r=2,3$ (matching Section 01's example exactly). $x(t)=c_1e^{2t}+c_2e^{3t}$ .

Reduce the system $\dfrac{dx}{dt}=4x-2y$ , $\dfrac{dy}{dt}=x+y$ to a single second-order equation in $x$ using elimination, and solve for $x(t)$ .

EXERCISE 19.6

Two coupled inventories where each one's growth rate depends on both current levels is exactly the elimination-method system setup from Section 04.

Two coupled asset positions follow $\dfrac{dx}{dt}=x+2y$ , $\dfrac{dy}{dt}=2x+y$ (matching Section 04's worked system exactly). From Section 04, $x(t)=c_1e^{3t}+c_2e^{-t}$ , $y(t)=c_1e^{3t}-c_2e^{-t}$ . As $t\to\infty$ , the $e^{3t}$ term dominates both, so both positions grow without bound at the same exponential rate, proportionally converging toward $x(t)\approx y(t)\approx c_1e^{3t}$ for large $t$ (assuming $c_1\neq0$ ) — the two positions become asymptotically equal in scale.

Two coupled asset positions follow $\dfrac{dx}{dt}=x+2y$ , $\dfrac{dy}{dt}=2x+y$ . Using Section 04's solution, describe the long-run behavior of $x(t)$ and $y(t)$ as $t\to\infty$ .

06 · Chapter Summary

Concept	Formula / Rule
Characteristic equation	$ar^2+br+c=0$ , from trying $y=e^{rx}$
Distinct real roots ( $\Delta>0$ )	$y=c_1e^{r_1x}+c_2e^{r_2x}$
Repeated root ( $\Delta=0$ )	$y=c_1e^{rx}+c_2xe^{rx}$
Complex roots ( $\Delta<0$ , $r=\alpha\pm\beta i$ )	$y=e^{\alpha x}(c_1\cos\beta x+c_2\sin\beta x)$
Nonhomogeneous	$y=y_h+y_p$
Systems via elimination	Differentiate one equation, substitute to remove the other unknown and its derivative, solve the resulting single equation

Next: Chapter 20 — Taylor Expansions and Approximation in Applied Contexts, which puts Chapters 10–11's series machinery to direct practical use.