Chapter 18

Medium

Differential Equations — First Order

00 · Symbol Glossary

$\dfrac{dy}{dx} = f(x,y)$first-order differential equation

An equation relating a function $y=y(x)$ to its own derivative. "First-order" means only the first derivative appears — no $y''$ or higher. Solving it means finding the function $y(x)$ itself, not just a number.

$C$arbitrary constant of a general solution

A free constant appearing in a general solution — the same role played by the constant of integration in Chapter 06, since solving a differential equation always involves at least one integration step.

$y(x_0)=y_0$initial condition

An extra piece of information — the value of $y$ at one specific point $x_0$ — used to pin down the constant $C$ and select one particular solution out of the whole general-solution family.

$\text{IVP}$initial value problem

A differential equation paired with an initial condition, together demanding a single, specific solution rather than a whole family of them.

$\mu(x)$mu of x — integrating factor

A helper function, introduced in Section 04, multiplied through a linear first-order equation to make its left side collapse into a single product-rule derivative, allowing direct integration.

01 · What a Differential Equation Is

Every derivative computed in Chapters 02–17 started from a known function and produced a new one. A differential equation runs this backward: it states a relationship the derivative must satisfy, and the task is to recover the original function.

Definition — Differential Equation, Order, General and Particular Solutions

A differential equation is an equation involving an unknown function and one or more of its derivatives. Its order is the highest derivative appearing in it.

A general solution is a family of functions (typically containing an arbitrary constant $C$ ) satisfying the equation. A particular solution is one specific member of that family, usually selected by an initial condition.

Example — Verifying a solution by substitution

Is $y=3e^{2x}$ a solution of $\dfrac{dy}{dx} = 2y$ ?

Differentiate the proposed $y$ (Chapter 02, Section 10, combined with the chain rule of Chapter 03): $\dfrac{dy}{dx} = 3\cdot2e^{2x} = 6e^{2x}$ .

Compare to $2y = 2(3e^{2x}) = 6e^{2x}$ . Since $\dfrac{dy}{dx}=2y$ holds for every $x$ , $y=3e^{2x}$ is a solution.

02 · Separable Equations

Many first-order equations can be rearranged so that every $x$ sits on one side and every $y$ on the other — at which point each side can be integrated separately using Chapters 06–07's tools.

Definition — Separable Equation

A first-order equation is separable if it can be written as $\dfrac{dy}{dx} = g(x)h(y)$ . Rearranging (treating $\dfrac{dy}{dx}$ formally as a ratio of differentials, in the spirit of Chapter 02's Leibniz notation):

\frac{1}{h(y)}\,dy = g(x)\,dx

Both sides can then be integrated independently.

Step-by-step — Solve $\dfrac{dy}{dx} = xy$

Confirm the equation is separable:

xy = x\cdot y

splits cleanly into a function of

x

alone (

x

) times a function of

y

alone (

y

Separate the variables: divide both sides by

y

(assuming

y\neq0

for now) and multiply by

dx

\dfrac{1}{y}\,dy = x\,dx

Integrate both sides separately (Chapter 07's log rule on the left, Chapter 06's power rule on the right):

\displaystyle\int\frac1y\,dy = \int x\,dx \implies \ln\lvert y\rvert = \frac{x^2}{2}+C_1

Solve for $y$ by exponentiating both sides (Chapter 02, Section 10's inverse relationship between $\ln$ and $e^x$ ):

\lvert y\rvert = e^{x^2/2+C_1} = e^{C_1}\cdot e^{x^2/2}

Rename the constant:

e^{C_1}

is just some positive number; absorbing the sign from

\lvert y\rvert

lets

C=\pm e^{C_1}

range over all nonzero reals:

y = Ce^{x^2/2}

❌ Losing the y = 0 solution when dividing by y

Step 2 above divided both sides by $y$ , silently assuming $y\neq0$ .

Why it breaks: $y=0$ (the constant zero function) also satisfies the original equation $\dfrac{dy}{dx}=xy$ , since $\dfrac{d}{dx}[0]=0$ and $x\cdot0=0$ for every $x$ — both sides equal $0$ . But $y=0$ can never be produced by $y=Ce^{x^2/2}$ for any constant $C$ , since $e^{x^2/2}>0$ always, so $Ce^{x^2/2}=0$ only if $C=0$ and the exponential vanishes — which it never does.

Consequence: dividing by an expression that could be zero can silently discard a valid solution. The complete solution set here is $y=Ce^{x^2/2}$ for any constant $C$ , including $C=0$ (which reproduces the lost $y=0$ solution as a limiting case) — in this particular equation the lost solution happens to be recoverable by allowing $C=0$ , but that must be checked, never assumed automatically.

03 · Initial Value Problems

A general solution is a whole family of curves. An initial condition selects exactly one.

Step-by-step — Solve the IVP $\dfrac{dy}{dx}=xy$, $y(0)=5$

Start from the general solution found in Section 02:

y=Ce^{x^2/2}

Substitute the initial condition $x=0$ , $y=5$ :

5 = Ce^{0^2/2} = Ce^0 = C(1) = C

Conclude $C=5$ , and write the particular solution:

y = 5e^{x^2/2}

04 · Linear First-Order Equations and the Integrating Factor

Not every first-order equation separates. A different, broader class — linear first-order equations — has its own dedicated method.

Definition — Linear First-Order Equation and Integrating Factor

A first-order equation is linear if it can be written as

\frac{dy}{dx} + P(x)y = Q(x)

Multiplying both sides by the integrating factor $\mu(x) = e^{\int P(x)\,dx}$ makes the left side collapse into a single derivative:

\frac{d}{dx}\big[\mu(x)y\big] = \mu(x)Q(x)

Why this particular μ(x) works

By the product rule (Chapter 02) and the chain rule (Chapter 03), $\dfrac{d}{dx}[\mu y] = \mu'y+\mu y' = \mu\left(\dfrac{\mu'}{\mu}\right)y + \mu y'$ . Since $\mu=e^{\int P\,dx}$ , differentiating gives $\mu' = P(x)\mu$ (Chapter 03's chain rule: differentiating the exponent $\int P\,dx$ brings down $P(x)$ by FTC Part 1, Chapter 06), so $\dfrac{\mu'}{\mu}=P(x)$ . Substituting: $\dfrac{d}{dx}[\mu y] = \mu P y + \mu y' = \mu(y'+Py)$ — exactly $\mu$ times the left side of the original equation, which is why multiplying through by $\mu$ turns the sum $y'+Py$ into the single collapsed derivative $\dfrac{d}{dx}[\mu y]$ .

Step-by-step — Solve $\dfrac{dy}{dx} + 2y = 6$

Identify $P(x)$ and $Q(x)$ : matching the standard form,

P(x)=2

(a constant),

Q(x)=6

Compute the integrating factor:

\mu(x) = e^{\int 2\,dx} = e^{2x}

(Chapter 06 power rule for integration, with the arbitrary constant of integration dropped, since any one antiderivative works here — the same freedom used with FTC Part 2 in Chapter 06).

Multiply both sides of the original equation by $\mu(x)=e^{2x}$ :

e^{2x}\dfrac{dy}{dx}+2e^{2x}y = 6e^{2x}

Recognize the left side as a collapsed derivative (Section 04's formula):

\dfrac{d}{dx}\big[e^{2x}y\big] = 6e^{2x}

Integrate both sides with respect to $x$ (Chapter 06, 07):

e^{2x}y = \displaystyle\int 6e^{2x}\,dx = 3e^{2x}+C

(u-substitution, Chapter 07, with

u=2x

Solve for $y$ by dividing through by $e^{2x}$ :

y = 3 + Ce^{-2x}

Verify:

\dfrac{dy}{dx} = -2Ce^{-2x}

. Check

\dfrac{dy}{dx}+2y = -2Ce^{-2x}+2(3+Ce^{-2x}) = -2Ce^{-2x}+6+2Ce^{-2x}=6

✓.

❌ Forgetting to multiply Q(x) by the integrating factor too

On the same equation, a common slip: multiplying the left side by $\mu(x)=e^{2x}$ correctly, but leaving the right side as plain $6$ instead of $6e^{2x}$ — writing $\dfrac{d}{dx}[e^{2x}y] = 6$ .

Why it breaks: the integrating factor is multiplied through the entire equation, both sides — it isn't a rewriting trick applied only to the left side. Skipping the right side breaks the equality: the left side genuinely equals $6e^{2x}$ after substitution (step 3), not $6$ .

Consequence: integrating the wrong right side gives $e^{2x}y = 6x+C \implies y = 6xe^{-2x}+Ce^{-2x}$ — plugging this back into the original equation does not satisfy $y'+2y=6$ (differentiating and simplifying leaves leftover $x$ -dependent terms that don't cancel), confirming the error. Every term on both sides must be multiplied by $\mu(x)$ .

05 · Application — Exponential Growth and Decay

Definition — Exponential Growth/Decay Equation

The separable equation $\dfrac{dy}{dt} = ky$ (rate of change proportional to current amount) has general solution $y = y_0e^{kt}$ , where $y_0=y(0)$ .

Step-by-step — Derive the solution and apply it to a decaying investment balance

Separate variables (Section 02's method):

\dfrac{1}{y}\,dy = k\,dt

Integrate both sides:

\ln\lvert y\rvert = kt+C_1 \implies y = Ce^{kt}

(same steps as Section 02's worked example, with

x^2/2

replaced by

kt

Apply the initial condition $y(0)=y_0$ :

y_0 = Ce^{0}=C

, so

y=y_0e^{kt}

Apply to a balance $B(t)$ starting at $10{,}000$ dollars, shrinking at continuous rate $k=-0.03$ (a $3\%$ continuous outflow):

B(t) = 10{,}000\,e^{-0.03t}

Evaluate after $5$ years:

B(5) = 10{,}000\,e^{-0.15} \approx 10{,}000(0.8607) \approx 8607

dollars.

06 · Exercises

EXERCISE 18.1

Separate the variables, integrate each side using Chapter 06/07 tools, then solve explicitly for $y$ .

$\dfrac{dy}{dx} = \dfrac{x}{y}$ . Separate: $y\,dy = x\,dx$ . Integrate: $\dfrac{y^2}{2}=\dfrac{x^2}{2}+C_1 \implies y^2 = x^2+C$ (renaming $2C_1$ as $C$ ). Solving explicitly: $y=\pm\sqrt{x^2+C}$ .

Solve the separable equation $\dfrac{dy}{dx} = \dfrac{x}{y}$ .

EXERCISE 18.2

Find the general solution first, then use the given point to solve for the constant.

$\dfrac{dy}{dx}=\dfrac{x}{y}$ , $y(0)=4$ . From Exercise 18.1, $y^2=x^2+C$ . At $x=0,y=4$ : $16=0+C \implies C=16$ . Particular solution: $y^2=x^2+16$ , i.e. $y=\sqrt{x^2+16}$ (taking the positive root to match $y(0)=4>0$ ).

Solve the initial value problem $\dfrac{dy}{dx}=\dfrac{x}{y}$ , $y(0)=4$ .

EXERCISE 18.3

Identify $P(x)$ and $Q(x)$ in the standard linear form, compute the integrating factor, then multiply through and integrate.

$\dfrac{dy}{dx} - 3y = 12$ : $P(x)=-3$ , $Q(x)=12$ . $\mu(x) = e^{\int -3\,dx} = e^{-3x}$ . Multiply through: $\dfrac{d}{dx}[e^{-3x}y] = 12e^{-3x}$ . Integrate: $e^{-3x}y = \dfrac{12}{-3}e^{-3x}+C = -4e^{-3x}+C$ . Solve: $y = -4+Ce^{3x}$ .

Solve $\dfrac{dy}{dx} - 3y = 12$ using an integrating factor.

EXERCISE 18.4

Determine which specific value of $y$ makes both sides of the original equation equal to zero, independent of any algebraic rearrangement.

For $\dfrac{dy}{dx}=xy$ (Section 02), $y=0$ satisfies the equation directly: $\dfrac{d}{dx}[0]=0$ and $x(0)=0$ , both sides match for every $x$ . Checking whether the general solution $y=Ce^{x^2/2}$ captures it: setting $C=0$ gives $y=0$ exactly. So the lost solution is recovered as the $C=0$ case here — but this recovery is not automatic in every separable equation and must always be checked separately.

For the equation $\dfrac{dy}{dx}=xy$ from Section 02, verify directly that $y=0$ is a solution, and check whether it is captured by the general solution family $y=Ce^{x^2/2}$ .

EXERCISE 18.5

This is the exponential growth/decay equation from Section 05 — identify $k$ from the given continuous rate and apply the solution formula directly.

A quantity grows according to $\dfrac{dQ}{dt}=0.08Q$ , $Q(0)=2000$ . By Section 05, $Q(t)=2000e^{0.08t}$ . At $t=10$ : $Q(10)=2000e^{0.8}\approx2000(2.2255)\approx4451$ .

A quantity satisfies $\dfrac{dQ}{dt}=0.08Q$ with $Q(0)=2000$ . Find $Q(10)$ .

EXERCISE 18.6

Recognize the equation as linear first-order in standard form, find the integrating factor, and solve; then apply the initial condition.

A bond's price $P(t)$ pays continuous coupon income while decaying toward face value: $\dfrac{dP}{dt} + 0.05P = 5$ , $P(0)=80$ . $\mu(t)=e^{0.05t}$ . $\dfrac{d}{dt}[e^{0.05t}P] = 5e^{0.05t}$ . Integrate: $e^{0.05t}P = \dfrac{5}{0.05}e^{0.05t}+C = 100e^{0.05t}+C$ . Solve: $P=100+Ce^{-0.05t}$ . Apply $P(0)=80$ : $80=100+C \implies C=-20$ . Particular solution: $P(t) = 100-20e^{-0.05t}$ — the price rises from $80$ toward a long-run value of $100$ as $t\to\infty$ .

A bond's price follows $\dfrac{dP}{dt}+0.05P=5$ with $P(0)=80$ . Solve for $P(t)$ and describe its long-run behavior as $t\to\infty$ .

07 · Chapter Summary

Concept	Formula / Rule
Order	Highest derivative appearing in the equation
General vs. particular solution	Family with constant $C$ vs. one member fixed by an initial condition
Separable equation	$\dfrac{dy}{dx}=g(x)h(y)$ ; integrate $\frac1h\,dy=g\,dx$ separately
Initial value problem	Equation + $y(x_0)=y_0$ ; solves for $C$
Linear first-order form	$y'+P(x)y=Q(x)$
Integrating factor	$\mu(x)=e^{\int P(x)\,dx}$ ; collapses left side to $\dfrac{d}{dx}[\mu y]$
Exponential growth/decay	$\dfrac{dy}{dt}=ky \implies y=y_0e^{kt}$

Next: Chapter 19 — Differential Equations: Second Order and Systems, which extends these methods to equations involving $y''$ and to multiple unknown functions changing together.