First- and Second-Order Differential Equations

Concise answer

Exam differential equations are won at the classification step. First order splits into separable (integrate both sides after separating) and linear (multiply by $e^{\int p}$ so the left side becomes a product derivative). Second order with constant coefficients is decided entirely by the characteristic roots, and the three root cases — distinct real, repeated, complex conjugate — carry three fixed solution shapes.

Definitions

Order: The order of the highest derivative appearing in the equation. A general solution of an $n$ th-order equation carries $n$ arbitrary constants.
Linear equation: An equation in which $y$ and its derivatives appear only to the first power and are not multiplied together, such as $y'' + p(x)y' + q(x)y = f(x)$ .
Separable equation: A first-order equation that can be written $\dfrac{dy}{dx} = g(x)h(y)$ , so that $\dfrac{dy}{h(y)} = g(x)\,dx$ can be integrated term by term.
Integrating factor: The multiplier $\mu(x) = e^{\int p(x)\,dx}$ that turns $y' + p(x)y$ into $\dfrac{d}{dx}\left[\mu(x)y\right] / \mu(x)$ , making the equation directly integrable.
Characteristic equation: The polynomial $ar^2 + br + c = 0$ obtained by substituting $y = e^{rx}$ into $ay'' + by' + cy = 0$ ; its roots determine the solution shape.
Initial-value problem: A differential equation together with values of $y$ (and, for second order, $y'$ ) at one point, which fixes the arbitrary constants.

Intuition

The integrating factor is not a trick to memorise but a reverse product rule. You want the left side of $y' + p y = q$ to be the derivative of something; multiplying by $\mu$ gives $\mu y' + \mu p y$ , which is $(\mu y)'$ precisely when $\mu' = p\mu$ — and that little equation is separable, with solution $\mu = e^{\int p}$ .

The characteristic equation appears because $e^{rx}$ turns differentiation into multiplication by $r$ . Complex roots are not a separate world: $e^{(\alpha + i\beta)x} = e^{\alpha x}(\cos \beta x + i \sin \beta x)$ , so the real and imaginary parts give the real solution pair, with $\alpha$ controlling growth or decay and $\beta$ controlling oscillation frequency.

Concept walkthrough

Classification comes first. Read off the order, then ask whether the equation is linear. A first-order equation that factors as $\dfrac{dy}{dx} = g(x)h(y)$ is separable: divide by $h(y)$ , multiply by $dx$ , integrate both sides, and remember that dividing by $h(y)$ can discard the constant solutions where $h(y) = 0$ . A first-order equation in the form $y' + p(x)y = q(x)$ is linear: multiply by $\mu(x) = e^{\int p\,dx}$ , recognise the left side as $(\mu y)'$ , integrate once, and divide by $\mu$ . Some equations are both, and either route works.

For $ay'' + by' + cy = 0$ with constant coefficients, substitute $y = e^{rx}$ and solve $ar^2 + br + c = 0$ . Distinct real roots $r_1 \neq r_2$ give $y = c_1 e^{r_1 x} + c_2 e^{r_2 x}$ . A repeated root $r$ gives $y = (c_1 + c_2 x)e^{rx}$ — the factor of $x$ is required because the two roots collapsed and a second independent solution is still needed. Complex conjugate roots $\alpha \pm \beta i$ give $y = e^{\alpha x}\left(c_1 \cos \beta x + c_2 \sin \beta x\right)$ . Every constant-coefficient homogeneous item on the exam is one of these three lines.

Nonhomogeneous equations use superposition: the general solution is $y = y_p + y_h$ , a particular solution plus the general homogeneous solution. Undetermined coefficients guesses $y_p$ with the same shape as the forcing term — a polynomial for a polynomial, $Ae^{kx}$ for an exponential, $A\cos \omega x + B \sin \omega x$ for a sinusoid — and always includes the full family, so a forcing term $e^{kx}$ needs the trial $Ae^{kx}$ and a forcing term $\sin \omega x$ needs both sine and cosine. If the trial function already solves the homogeneous equation, multiply it by $x$ (or $x^2$ for a repeated root) before solving for the coefficients.

After this page, you should be able to

Classify an equation by order and linearity, and choose separation or an integrating factor for a first-order problem.
Solve a first-order linear equation with the integrating factor and impose an initial condition.
Write the general solution of $ay'' + by' + cy = 0$ from the characteristic roots in all three cases.
Recognise a repeated root and supply the missing $x e^{rx}$ solution rather than a second copy of $e^{rx}$ .
Set up a particular solution by undetermined coefficients and detect when resonance forces an extra factor of $x$ .

Formulas and assumptions

Integrating factor for a first-order linear equation

\mu(x) = e^{\int p(x)\,dx}, \qquad y = \frac{1}{\mu(x)}\left( \int \mu(x) q(x)\,dx + C \right)

Variables

p(x): the coefficient of y after the equation is written in standard form
q(x): the forcing term
C: the constant of integration

Assumptions

The equation is first written in standard form with coefficient 1 on y'.
p and q are continuous on the interval of interest.

Separable equation

\frac{dy}{dx} = g(x)h(y) \;\Longrightarrow\; \int \frac{dy}{h(y)} = \int g(x)\,dx

Variables

g(x): the x-only factor
h(y): the y-only factor

Assumptions

h(y) is nonzero on the interval used; each root of h gives a constant solution that separation drops.

Characteristic equation, three cases

ar^{2} + br + c = 0 \;\Rightarrow\; y = \begin{cases} c_1 e^{r_1 x} + c_2 e^{r_2 x}, & r_1 \neq r_2 \text{ real} \\ (c_1 + c_2 x)e^{rx}, & \text{repeated } r \\ e^{\alpha x}\left(c_1 \cos \beta x + c_2 \sin \beta x\right), & r = \alpha \pm \beta i \end{cases}

Variables

a, b, c: the constant coefficients
r: a root of the characteristic polynomial
alpha, beta: the real and imaginary parts of a complex root

Assumptions

The coefficients are constants and a is nonzero.
The equation is homogeneous; a forcing term needs a particular solution added.

Worked example

A first-order linear initial-value problem

Solve $\displaystyle y' + \frac{2}{x}\,y = x$ for $x > 0$ with $y(1) = 1$ .

1The equation is already in standard form $y' + p(x)y = q(x)$ with $p(x) = \dfrac{2}{x}$ and $q(x) = x$ , so it is first-order linear and the integrating factor applies.
2Compute $\mu(x) = e^{\int (2/x)\,dx} = e^{2\ln x} = x^{2}$ for $x > 0$ .
3Multiply through: $x^{2}y' + 2xy = x^{3}$ , and the left side is exactly $\dfrac{d}{dx}\left(x^{2}y\right)$ .
4Integrate: $x^{2}y = \dfrac{x^{4}}{4} + C$ , so $y = \dfrac{x^{2}}{4} + \dfrac{C}{x^{2}}$ .
5Impose $y(1) = 1$ : $\dfrac{1}{4} + C = 1$ , so $C = \dfrac{3}{4}$ .
6Check by substitution: with $y = \dfrac{x^{2}}{4} + \dfrac{3}{4x^{2}}$ we get $y' = \dfrac{x}{2} - \dfrac{3}{2x^{3}}$ , and $y' + \dfrac{2}{x}y = \dfrac{x}{2} - \dfrac{3}{2x^{3}} + \dfrac{x}{2} + \dfrac{3}{2x^{3}} = x$ .

$y = \dfrac{x^{2}}{4} + \dfrac{3}{4x^{2}}$ for $x > 0$ .

Common traps

Forgetting to divide by the leading coefficient before reading off $p(x)$ ; the integrating factor is built from the standard form only.
Writing a repeated root as $c_1 e^{rx} + c_2 e^{rx}$ , which is a single solution with a renamed constant — the second independent solution is $x e^{rx}$ .
Mishandling a negative discriminant: $\sqrt{-36}$ is $6i$ , so the roots of $r^2 - 4r + 13$ are $2 \pm 3i$ , not $2 \pm 3$ .
Losing constant solutions when separating: dividing by $h(y)$ silently discards every $y$ with $h(y) = 0$ .
Guessing only $\sin \omega x$ for a sinusoidal forcing term, or omitting the extra factor of $x$ when the trial solution already solves the homogeneous equation.

A Second-Order Initial-Value Problem with Complex Roots GRE Mathematics overview Limits and Continuity Eigenvalues and Determinants Sign in for current practice

Question depth and domain coverage vary by exam. Practice answers are checked after submission.

Sources

GRE Subject Test Content and Structure — ETS. Accessed 2026-07-06. Use as a cited source for exam facts; do not imply affiliation or reproduce protected test material.
Calculus Volume 2, Section 4.1: Basics of Differential Equations — OpenStax. Accessed 2026-08-15. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
Calculus Volume 2, Section 4.3: Separable Equations — OpenStax. Accessed 2026-08-15. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
Calculus Volume 2, Section 4.5: First-order Linear Equations — OpenStax. Accessed 2026-08-15. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
Calculus Volume 3, Section 7.1: Second-Order Linear Equations — OpenStax. Accessed 2026-08-15. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
Calculus Volume 3, Section 7.2: Nonhomogeneous Linear Equations — OpenStax. Accessed 2026-08-15. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.

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First- and Second-Order Differential Equations

Concise answer

Definitions

Intuition

Concept walkthrough

After this page, you should be able to

Formulas and assumptions

Integrating factor for a first-order linear equation

Separable equation

Characteristic equation, three cases

Worked example

A first-order linear initial-value problem

Common traps

Related pages and practice

Sources

Sources and corrections