ODE Terminology

The Language of
Solutions

Every term you will ever need — general, particular, complementary, explicit, implicit, IVP — built from first principles with full rigor.

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What Does "Solution" Mean?

A solution is a function \(\phi(x)\) that, when substituted into the ODE — including every derivative — reduces it to an identity that holds for all \(x\) in the domain.

Definition — Solution of an ODE
\(\phi\) defined on \(I\) is a solution of \(F(x,y,y',\ldots,y^{(n)})=0\) if it is \(n\)-times differentiable and \[F\!\left(x,\,\phi(x),\,\phi'(x),\,\ldots,\,\phi^{(n)}(x)\right) = 0 \qquad \forall\, x \in I.\]
Intuition
The ODE is a lock. A solution is any key shape that fits. Usually not just one key — a whole family of correctly shaped keys exists. Initial conditions specify which key you actually want.

Verification by substitution

Is \(y = Ce^{2x}\) a solution of \(y' - 2y = 0\)?

Start with the given function
\[y = Ce^{2x}\] \[y' = 2Ce^{2x}\] \[y' - 2y = 2Ce^{2x} - 2(Ce^{2x})\] \[y' - 2y = 0 \quad \checkmark\]

True for every value of \(C\). So this is a one-parameter family of solutions.

General vs Particular Solution

General Solution
\(y = c_1 e^x + c_2 e^{2x}\)

The entire family. Contains \(n\) arbitrary constants for an \(n\)-th order ODE.

Particular Solution
\(y = 3e^x - e^{2x}\)

One specific member of the family. All constants are fixed — no free parameters remain.

Memory Anchor
General solution = the whole squad. Particular solution = one specific member of the squad, picked by applying conditions.

For \(y'' - 3y' + 2y = 0\):

Characteristic equation
\[r^2 - 3r + 2 = 0\] \[(r - 1)(r - 2) = 0\] \[r = 1, \quad r = 2\]
General solution (2-parameter family)
\[y = c_1 e^x + c_2 e^{2x}\]
Differentiate the general solution
\[y = c_1 e^x + c_2 e^{2x}\] \[y' = c_1 e^x + 2c_2 e^{2x}\]
Apply \(y(0) = 1\)
\[c_1 e^0 + c_2 e^0 = 1\] \[c_1 + c_2 = 1 \quad \cdots (1)\]
Apply \(y'(0) = 0\)
\[c_1 e^0 + 2c_2 e^0 = 0\] \[c_1 + 2c_2 = 0 \quad \cdots (2)\]
Subtract (1) from (2)
\[(c_1 + 2c_2) - (c_1 + c_2) = 0 - 1\] \[c_2 = -1\]
Back-substitute into (1)
\[c_1 + (-1) = 1\] \[c_1 = 2\]
Particular solution
\[y = 2e^x - e^{2x}\]

Homogeneous vs Nonhomogeneous

Homogeneous
\(Ly = 0\)

RHS is zero. No external forcing. Describes free, undriven behavior.

Nonhomogeneous
\(Ly = g(x) \neq 0\)

Driven by an external input \(g(x)\). Needs two pieces to solve completely.

Physical Intuition
A spring with no hand pushing it (\(g=0\)) just oscillates on its own — homogeneous. Push and pull it rhythmically (\(g\neq 0\)) — nonhomogeneous. The ODE records the difference.
Same operator, two equations
\[y'' - 3y' + 2y = 0 \quad \text{homogeneous — free oscillation}\] \[y'' - 3y' + 2y = e^x \quad \text{nonhomogeneous — forced by } e^x\]

Complementary Function \(y_c\) and Particular Solution \(y_p\)

Complementary Function \(y_c\)
\(Ly_c = 0\)

General solution of the associated homogeneous equation. Has free constants. Encodes all natural, unforced behavior.

Particular Solution \(y_p\)
\(Ly_p = g(x)\)

One specific solution of the full forced equation. No free constants. Encodes forced response only.

Do Not Conflate
\(y_c\) is the general solution of \(Ly=0\) — it does NOT satisfy the nonhomogeneous equation. \(y_p\) is one solution of \(Ly=g(x)\) — it has no free constants. They are structurally different objects that are added together to form the complete solution.

Why must \(y_c\) exist as a separate piece?

Let \(y_1\) and \(y_2\) both satisfy \(Ly = g(x)\). By linearity of \(L\):

Linearity argument
\[Ly_1 = g(x)\] \[Ly_2 = g(x)\] \[L(y_1 - y_2) = Ly_1 - Ly_2\] \[L(y_1 - y_2) = g(x) - g(x)\] \[L(y_1 - y_2) = 0\]

Every difference between two particular solutions satisfies the homogeneous equation. The full set of such differences is exactly \(y_c\). So \(y_p\) anchors you to one correct answer, and \(y_c\) encodes all the remaining freedom. That is why the complete solution must be \(y_c + y_p\).

The characteristic roots are \(r=1,2\). Since \(r=1\) matches the forcing term \(e^x\), the naive guess \(Ae^x\) duplicates \(y_c\) and produces zero when substituted. Multiply by \(x\):

Ansatz (resonance — multiply by x)
\[y_p = Axe^x\] \[y_p' = Ae^x + Axe^x\] \[y_p' = A(1 + x)e^x\] \[y_p'' = Ae^x + A(1+x)e^x\] \[y_p'' = A(2 + x)e^x\]
Substitute into \(y'' - 3y' + 2y = e^x\)
\[A(2+x)e^x - 3A(1+x)e^x + 2Axe^x = e^x\] \[e^x\!\left[A(2+x) - 3A(1+x) + 2Ax\right] = e^x\] \[e^x\!\left[(2A - 3A) + (A - 3A + 2A)x\right] = e^x\] \[e^x\!\left[-A + 0\cdot x\right] = e^x\] \[-A = 1\] \[A = -1\]
Particular solution
\[y_p = -xe^x\]

The Master Formula: \(y = y_c + y_p\)

Theorem — General Solution of a Nonhomogeneous Linear ODE
If \(y_p\) is any particular solution of \(Ly = g(x)\) and \(y_c\) is the general solution of \(Ly = 0\), then the general solution of \(Ly = g(x)\) is \[\boxed{y = y_c + y_p.}\]

Proof — both directions, stacked

Forward: any solution y of Ly = g has the form yc + yp
\[Ly = g(x)\] \[Ly_p = g(x)\] \[L(y - y_p) = Ly - Ly_p\] \[L(y - y_p) = g(x) - g(x)\] \[L(y - y_p) = 0\] \[y - y_p \;\in\; \{y_c\}\] \[y = y_c + y_p \quad \checkmark\]
Backward: any yc + yp satisfies Ly = g
\[L(y_c + y_p) = Ly_c + Ly_p\] \[L(y_c + y_p) = 0 + g(x)\] \[L(y_c + y_p) = g(x) \quad \checkmark\]
Physical Meaning
\(y_c\) = natural, free behavior (what the system does on its own). \(y_p\) = forced response (what the driving input causes). Total output = their sum. This is the superposition principle for linear systems.

Full general solution for \(y'' - 3y' + 2y = e^x\):

\[y_c = c_1 e^x + c_2 e^{2x}\] \[y_p = -xe^x\] \[y = c_1 e^x + c_2 e^{2x} - xe^x\]

Explicit vs Implicit Solutions

Explicit Solution
\(y = f(x)\)

\(y\) is isolated on the left. Direct and evaluatable.

Implicit Solution
\(G(x,y) = 0\)

\(y\) lives inside a relation with \(x\). Not isolated. Still a genuine solution.

Warning
An implicit solution still satisfies the ODE when differentiated implicitly. Many real ODEs yield implicit solutions that cannot be algebraically isolated. They are no less valid for it.

Example — separable ODE giving an implicit solution

Solve \(dy/dx = -x/y\).

Separate variables
\[y\,dy = -x\,dx\]
Integrate both sides
\[\int y\,dy = -\int x\,dx\] \[\frac{y^2}{2} = -\frac{x^2}{2} + C\]
Implicit general solution (family of circles)
\[x^2 + y^2 = K, \quad K = 2C > 0\]
Explicit particular solution — upper branch, K = 25
\[y = \sqrt{25 - x^2}\]

The implicit form captures both branches at once. The explicit form isolates one.

Initial-Value Problem (IVP)

An IVP is an ODE together with \(n\) initial conditions that fix \(y\) and its first \(n-1\) derivatives at a single point \(x_0\). Each condition gives one scalar equation in the constants, forming a square system that uniquely determines all \(n\) free constants.

Definition — IVP (\(n\)-th order)
Solve \(F(x,y,y',\ldots,y^{(n)}) = 0\) subject to \[y(x_0) = y_0,\quad y'(x_0) = y_1,\quad \ldots,\quad y^{(n-1)}(x_0) = y_{n-1}.\]

Interactive IVP Solver

For \(y'' - y = 0\), the general solution is \(y = c_1 e^x + c_2 e^{-x}\). Enter initial conditions to pin down \(c_1\) and \(c_2\):

Enter values and click Solve.

Full Worked Example: All Terms in One Problem

The problem
\[y'' - y = 2, \qquad y(0) = 1,\quad y'(0) = 0\]
1
Write the associated homogeneous equation.
Set RHS = 0
\[y'' - y = 0\]
// This is the homogeneous equation.
2
Solve for \(y_c\) — the complementary function.
Characteristic equation
\[r^2 - 1 = 0\] \[(r - 1)(r + 1) = 0\] \[r = 1, \quad r = -1\]
Complementary function
\[y_c = c_1 e^x + c_2 e^{-x}\]
// General solution of the homogeneous part. Has free constants.
3
Find \(y_p\) — one solution of the full equation.
Guess constant (since g = 2 is constant)
\[y_p = A\] \[y_p' = 0\] \[y_p'' = 0\] \[y_p'' - y_p = 0 - A\] \[-A = 2\] \[A = -2\] \[y_p = -2\]
// One specific function satisfying y'' - y = 2. No free constants.
4
Write the general solution of the nonhomogeneous ODE.
\[y = y_c + y_p\] \[y = c_1 e^x + c_2 e^{-x} - 2\]
// Entire family. Two free constants remain.
5
Apply IVP conditions to find \(c_1\) and \(c_2\).
Differentiate
\[y' = c_1 e^x - c_2 e^{-x}\]
Apply y(0) = 1
\[c_1 e^0 + c_2 e^0 - 2 = 1\] \[c_1 + c_2 = 3 \quad \cdots (1)\]
Apply y'(0) = 0
\[c_1 e^0 - c_2 e^0 = 0\] \[c_1 - c_2 = 0 \quad \cdots (2)\]
Add (1) and (2)
\[2c_1 = 3\] \[c_1 = \tfrac{3}{2}\]
Subtract (2) from (1)
\[2c_2 = 3\] \[c_2 = \tfrac{3}{2}\]
6
Particular solution of the IVP — explicit form.
\[y = \tfrac{3}{2}e^x + \tfrac{3}{2}e^{-x} - 2\] \[y = 3\cosh(x) - 2\]
// One explicit function. No free constants. Unique answer.
Common Pitfall
The general solution of a nonhomogeneous ODE requires both \(y_c\) and \(y_p\). Writing only \(y_c\) solves the wrong (homogeneous) equation. Writing only \(y_p\) gives one valid answer but discards the entire family.

How All the Terms Relate

Every term is a row. Read top to bottom — each row produces or feeds into the next.

ODE \(F(x,y,y',\ldots)=0\) The raw equation. Any function that satisfies it identically is a solution.
set RHS=0  /  keep g(x)≠0
Homogeneous \(Ly = 0\) No forcing. Free, natural behavior only.
Nonhomogeneous \(Ly = g(x)\) Has a driving term. Forced behavior. Requires two pieces to solve.
solve Ly = 0
General Solution \(y = c_1\phi_1 + \cdots + c_n\phi_n\) Whole family of solutions to the homogeneous ODE. \(n\) free constants. Also serves as \(y_c\) in the nonhomogeneous setup.
re-labeled as yc / find yp separately
Complementary \(y_c\) \(Ly_c = 0\) General solution of the homogeneous part. Has free constants. Natural motion.
Particular \(y_p\) \(Ly_p = g(x)\) One specific solution of the forced equation. No constants. Forced response.
add together
Master Formula \(y = y_c + y_p\) General solution of the nonhomogeneous ODE. Every solution has this form.
apply initial conditions
IVP \(y(x_0)=y_0,\; y'(x_0)=y_1\) \(n\) conditions fix all \(n\) constants. Selects one particular solution.
result expressed as
Explicit \(y = f(x)\) \(y\) is isolated. Clean and directly evaluatable.
Implicit \(G(x,y) = 0\) \(y\) not isolated. Lives inside a relation. Still a valid solution.

Real-World EE: Series RLC Circuit

The series RLC circuit is a direct physical instantiation of a second-order nonhomogeneous linear ODE. Every vocabulary term on this page maps to something you can probe on a bench with an oscilloscope.

Physical Setup
Resistor \(R\), inductor \(L\), and capacitor \(C\) connected in series with a voltage source \(v_s(t)\). The unknown is \(q(t)\), the charge on the capacitor. Current is \(i = \dot{q}\). Kirchhoff's Voltage Law (KVL) gives the governing ODE directly.

Derive the ODE from KVL — stacked

KVL: sum of voltage drops equals source voltage
\[v_R + v_L + v_C = v_s(t)\]
Constitutive relations for each element
\[v_R = Ri = R\dot{q}\] \[v_L = L\dot{i} = L\ddot{q}\] \[v_C = \frac{q}{C}\]
Substitute into KVL
\[R\dot{q} + L\ddot{q} + \frac{q}{C} = v_s(t)\]
Rewrite in standard form
\[L\ddot{q} + R\dot{q} + \frac{1}{C}q = v_s(t)\]

This is the ODE. Every term maps to a vocabulary word from this page:

The ODE
\(L\ddot{q} + R\dot{q} + \tfrac{1}{C}q = v_s(t)\)

Governing equation. Unknown is \(q(t)\) — charge on the capacitor as a function of time.

Homogeneous version
\(L\ddot{q} + R\dot{q} + \tfrac{1}{C}q = 0\)

Circuit with no source. Describes how stored energy dissipates on its own — the natural ring-down.

Complementary function \(q_c\)
\(q_c = e^{-\alpha t}(c_1\cos\omega_d t + c_2\sin\omega_d t)\)

Transient response. Decays to zero as \(t\to\infty\) when \(R>0\). Contains free constants \(c_1, c_2\).

Particular solution \(q_p\)
\(q_p = Q_0\cos(\omega t - \phi)\)

Steady-state response. Oscillates at the source frequency forever. No free constants.

General solution
\(q = q_c + q_p\)

Total response = transient + steady-state. Designers care about \(q_p\) — the transient eventually dies.

Initial conditions (IVP)
\(q(0)=Q_0,\quad \dot{q}(0)=I_0\)

Initial charge on the capacitor and initial current. Fix \(c_1, c_2\) and give the unique circuit response.

Full Worked Example — Underdamped RLC, sinusoidal source

Values: \(L = 1\,\text{H}\), \(R = 2\,\Omega\), \(C = \tfrac{1}{2}\,\text{F}\), source \(v_s(t) = 10\cos(2t)\,\text{V}\). Initial state: \(q(0) = 0\), \(\dot{q}(0) = 0\) (initially uncharged, no current).

Substitute values into the governing ODE
\[(1)\ddot{q} + (2)\dot{q} + \frac{1}{1/2}q = 10\cos(2t)\] \[\ddot{q} + 2\dot{q} + 2q = 10\cos(2t)\]
1
Solve the homogeneous equation — transient (natural) response.
Set RHS = 0
\[\ddot{q} + 2\dot{q} + 2q = 0\]
Characteristic equation
\[r^2 + 2r + 2 = 0\] \[r = \frac{-2 \pm \sqrt{4 - 8}}{2}\] \[r = \frac{-2 \pm \sqrt{-4}}{2}\] \[r = \frac{-2 \pm 2i}{2}\] \[r = -1 \pm i\]
Complex roots — underdamped: \(\alpha = 1\), \(\omega_d = 1\)
\[q_c = e^{-t}(c_1\cos t + c_2\sin t)\]
// Transient response. Envelope decays as e^{-t} — dies off exponentially.
2
Find \(q_p\) — forced (steady-state) response.

Driving frequency \(\omega=2\). Natural frequency \(\omega_d=1\neq 2\), so no resonance — standard undetermined coefficients applies.

Ansatz
\[q_p = A\cos(2t) + B\sin(2t)\] \[\dot{q}_p = -2A\sin(2t) + 2B\cos(2t)\] \[\ddot{q}_p = -4A\cos(2t) - 4B\sin(2t)\]
Substitute into the ODE
\[\bigl(-4A\cos 2t - 4B\sin 2t\bigr) + 2\bigl(-2A\sin 2t + 2B\cos 2t\bigr) + 2\bigl(A\cos 2t + B\sin 2t\bigr) = 10\cos 2t\]
Collect cos(2t) terms
\[-4A + 4B + 2A = 10\] \[-2A + 4B = 10 \quad \cdots (1)\]
Collect sin(2t) terms
\[-4B - 4A + 2B = 0\] \[-4A - 2B = 0\] \[B = -2A \quad \cdots (2)\]
Substitute (2) into (1)
\[-2A + 4(-2A) = 10\] \[-2A - 8A = 10\] \[-10A = 10\] \[A = -1\] \[B = -2(-1) = 2\]
Particular solution
\[q_p = -\cos(2t) + 2\sin(2t)\]
// Permanent oscillation at source frequency 2 rad/s. This is the steady-state response.
3
General solution of the nonhomogeneous ODE.
\[q = q_c + q_p\] \[q = e^{-t}(c_1\cos t + c_2\sin t) - \cos(2t) + 2\sin(2t)\]
// Transient + steady-state. Two free constants c1, c2 remain.
4
Apply IVP: \(q(0)=0\), \(\dot{q}(0)=0\).
Apply q(0) = 0
\[e^0(c_1\cos 0 + c_2\sin 0) - \cos 0 + 2\sin 0 = 0\] \[c_1 \cdot 1 - 1 + 0 = 0\] \[c_1 = 1\]
Differentiate q
\[\dot{q} = e^{-t}\bigl[(-c_1+c_2)\cos t + (-c_2-c_1)\sin t\bigr] + 2\sin(2t) + 4\cos(2t)\]
Apply \(\dot{q}(0) = 0\)
\[e^0\bigl[(-c_1+c_2)\cos 0 + (-c_2-c_1)\sin 0\bigr] + 0 + 4 = 0\] \[(-c_1 + c_2)(1) + 0 + 4 = 0\] \[-c_1 + c_2 + 4 = 0\] \[-1 + c_2 + 4 = 0\] \[c_2 = -3\]
5
Particular solution of the IVP — the actual circuit behavior.
\[q(t) = e^{-t}(\cos t - 3\sin t) - \cos(2t) + 2\sin(2t)\]
// Explicit, fully determined. As t → ∞ the first term vanishes. What remains forever is qp.
Physical Interpretation — ODE vocabulary maps exactly to EE vocabulary

\(q_c = e^{-t}(\cos t - 3\sin t)\)  =  transient response — set by initial stored energy, decays to zero.

\(q_p = -\cos(2t) + 2\sin(2t)\)  =  steady-state response — oscillates at the source frequency indefinitely.

Sum = the full particular solution of the IVP. What the oscilloscope shows early in time includes both; after the transient dies, only \(q_p\) remains.

Resonance — When the Ansatz Must Be Modified
If the source frequency \(\omega\) equals the natural frequency \(\omega_0 = 1/\sqrt{LC}\), the standard guess \(q_p = A\cos(\omega_0 t)+B\sin(\omega_0 t)\) fails because those terms already appear in \(q_c\). Substituting them into the ODE produces \(0 = v_s\) — a contradiction. The fix is to multiply by \(t\):
\[q_p = t\bigl(A\cos(\omega_0 t) + B\sin(\omega_0 t)\bigr)\]
This is the same algebraic resonance signal that forced \(y_p = Axe^x\) instead of \(y_p = Ae^x\) earlier. In an undamped circuit (\(R=0\)) it produces amplitude that grows linearly without bound — physical resonance. In practice \(R>0\) limits growth, but near-resonance still causes large transient overshoots — a critical design concern.

Quick Reference Battle Card

general Solution
\(y = c_1\phi_1 + \cdots + c_n\phi_n\)

Entire family. \(n\) free constants for order \(n\).

particular Solution
\(y = 2e^x - e^{2x}\)

One member of the family. All constants fixed — usually by IVP.

complementary \(y_c\)
\(Ly_c = 0\)

General solution of \(Ly=0\). Natural behavior. Has free constants.

particular \(y_p\)
\(Ly_p = g(x)\)

One solution of the forced equation. Forced response only. No constants.

master formula
\(y = y_c + y_p\)

General solution of the nonhomogeneous ODE. Every solution has this form.

explicit
\(y = f(x)\)

\(y\) isolated. Directly evaluatable.

implicit
\(G(x,y) = 0\)

\(y\) not isolated. A relation. Still valid.

IVP
\(y(x_0)=y_0,\; y'(x_0)=y_1\)

Fixes all constants. Selects exactly one particular solution.