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Linear differential equations

⌚ ~59 min ▢ 7 blocks ✓ 27 exercises ➣ Prerequisites : Primitives --- practical techniques, Differentiability

A linear differential equation is an equation of the form $T(y) = f$ where the unknown is a function $y$, $T$ is a linear operator built from $y$ and its derivatives, and $f$ is a fixed function. This chapter treats two cases, the only ones in the program:

first-order : $y'(x) + a(x) \, y(x) = b(x)$, with $a, b \in C(I, \mathbb{K})$ given continuous functions on an interval $I$;
second-order with constant coefficients : $y''(x) + a \, y'(x) + b \, y(x) = f(x)$, with $a, b \in \mathbb{K}$ scalars and $f \in C(I, \mathbb{K})$.

Here $\mathbb{K} = \mathbb R$ or $\mathbb C$ and $I$ is a non-empty interval of $\mathbb R$ not reduced to a point.
The linearity of the operator $y \mapsto y' + a y$ (resp.\ $y \mapsto y'' + a y' + b y$) drives the whole structure of the chapter. Two consequences are recurring : the superposition principle (linear combinations of solutions of separate equations solve the combined equation) and the affine structure of the solution set (the set of solutions is the sum of a particular solution and the vector space of solutions of the associated homogeneous equation $T(y) = 0$).
The chapter has two sections. The first treats first-order linear ODEs : the homogeneous case via the integrating factor $e^A$ where $A$ is a primitive of $a$, then the complete case via variation of constants, then the Cauchy problem (existence + uniqueness, proof included), then an optional enrichment on raccord problems where the leading coefficient may vanish. The second section treats second-order constant-coefficient ODEs : the homogeneous case via the characteristic polynomial (3 cases by discriminant when $\mathbb{K} = \mathbb R$), then the complete case with the Cauchy theorem (statement admitted, proof out of program), then the practical ansatz for the program's three special right-hand sides (polynomial, $\alpha e^{\lambda x}$, $\beta \cos(\omega x) / \beta \sin(\omega x)$).
Three reflexes the reader should leave with : (i) the linear structure of the solution set is the central guiding idea --- every result is a consequence of the linearity of $y \mapsto y' + a y$ or $y \mapsto y'' + a y' + b y$ ; (ii) first-order solutions reduce to antidifferentiation (one primitive in the homogeneous case, two in the variation-of-constants formula), invoking the toolbox of chapter Primitives ; (iii) second-order solutions reduce to a characteristic polynomial (for the homogeneous part) and to an ansatz with coefficient identification (for a particular solution).

I First-order linear ODEs

I.1 Definition and homogeneous case

A linear first-order ODE has the form $y'(x) + a(x) \, y(x) = b(x)$, where $a, b$ are given continuous functions on $I$ with values in $\mathbb{K}$. Read literally, $y$ denotes an unknown function on $I$ ; we seek functions $y \in C^1(I, \mathbb{K})$ satisfying the equation pointwise on $I$. The equation is linear because the map $y \mapsto y' + a y$ is linear on $C^1(I, \mathbb{K})$ ; this drives the entire chapter's structural reasoning. The simplest case --- the homogeneous equation $y' + a(x) y = 0$ --- admits a one-line solution via the integrating factor $e^A$ where $A$ is a primitive of $a$ on $I$.

Definition — Linear first-order ODE

Let $I$ be an interval of $\mathbb R$ and $a, b \in C(I, \mathbb{K})$, where $\mathbb{K} = \mathbb R$ or $\mathbb C$. The linear first-order ODE associated with $a$ and $b$ is the equation $$ y'(x) + a(x) \, y(x) \ = \ b(x), \qquad x \in I, $$ of unknown $y \in C^1(I, \mathbb{K})$. The associated homogeneous equation is $y'(x) + a(x) \, y(x) = 0$. A solution on $I$ is a function $y \in C^1(I, \mathbb{K})$ satisfying the equation pointwise. We write $S$ for the solution set of the complete equation, $S_0$ for the solution set of the homogeneous equation.

Theorem — Solutions of the homogeneous equation $y' + a(x) y \equal 0$

Let $a \in C(I, \mathbb{K})$ and $A$ a primitive of $a$ on $I$. The solutions of the homogeneous equation $y'(x) + a(x) \, y(x) = 0$ on $I$ are exactly the functions $$ x \ \longmapsto \ \lambda \, e^{-A(x)}, \qquad \lambda \in \mathbb{K}. $$ Equivalently, $S_0 = \mathbb{K} \cdot e^{-A}$ is a one-dimensional $\mathbb{K}$-vector subspace of $C^1(I, \mathbb{K})$.

Proof

The key observation is that for any $y \in C^1(I, \mathbb{K})$, $$ \begin{aligned} \bigl( y(x) \, e^{A(x)} \bigr)' \ &= \ y'(x) \, e^{A(x)} + y(x) \, A'(x) \, e^{A(x)} && \text{(product rule)} \\ &= \ y'(x) \, e^{A(x)} + a(x) \, y(x) \, e^{A(x)} && \text{(since $A' = a$ on $I$)} \\ &= \ \bigl( y'(x) + a(x) \, y(x) \bigr) \, e^{A(x)} && \text{(factor $e^A$)}. \end{aligned} $$ Since $e^A$ never vanishes, $$ y'(x) + a(x) \, y(x) \ = \ 0 \quad \Longleftrightarrow \quad \bigl( y \, e^A \bigr)' \ = \ 0 \quad \Longleftrightarrow \quad y \, e^A \text{ is constant on } I. $$ We use here the basic fact that a function with zero derivative on an interval is constant : this is the uniqueness statement for primitives (already invoked in chapter Primitives) and is admitted at this stage ; the formal mean-value-theorem-based proof is consolidated in chapter Differentiability. So $y' + a y = 0$ if and only if there is $\lambda \in \mathbb{K}$ with $y(x) e^{A(x)} = \lambda$ for all $x \in I$, i.e.\ $y(x) = \lambda e^{-A(x)}$. Conversely, every function $y(x) = \lambda e^{-A(x)}$ is differentiable on $I$ with $$ y'(x) \ = \ - a(x) \, \lambda \, e^{-A(x)} \ = \ - a(x) \, y(x), $$ hence satisfies $y' + a y = 0$.

Method — Solving the homogeneous equation $y' + a(x) y \equal 0$

To find every solution of $y' + a(x) y = 0$ on an interval $I$ :

Compute a primitive $A$ of $a$ on $I$ (one antidifferentiation, using the table of chapter Primitives).
Write the general solution as $x \mapsto \lambda \, e^{-A(x)}$, with $\lambda \in \mathbb{K}$ a free parameter.

The solution set $S_0$ is a one-dimensional $\mathbb{K}$-vector subspace ; one degree of freedom, captured by the constant $\lambda$.

Example — Constant coefficient $a$

For $a \in \mathbb{K}$ constant, the equation $y' + a y = 0$ on $\mathbb R$ has $A(x) = a x$ as a primitive of the constant function $a$. Hence the solutions are $$ x \ \longmapsto \ \lambda \, e^{-a x}, \qquad \lambda \in \mathbb{K}. $$ In particular, with $a = -1$ and the initial condition $y(0) = 1$, this recovers the characterisation : the exponential is the unique $C^1$ function on $\mathbb R$ satisfying $y' = y$ and $y(0) = 1$.

Example — Variable coefficient

On $\mathbb R$, the equation $y' = \dfrac{y}{1 + x^2}$ rewrites as $y' - \dfrac{1}{1 + x^2} y = 0$. A primitive of $x \mapsto 1/(1 + x^2)$ is $\arctan$ (table from chapter Primitives), so $A(x) = -\arctan x$ is a primitive of $a(x) = -1/(1 + x^2)$. The general solution is $$ x \ \longmapsto \ \lambda \, e^{\arctan x}, \qquad \lambda \in \mathbb R. $$

Skills to practice

Solving homogeneous first-order ODEs

I.2 Complete equation$\virgule$ superposition$\virgule$ variation of constants

Once the homogeneous case is settled, the linearity of the map $y \mapsto y' + a y$ gives two structural results. The first --- the superposition principle --- says that solutions of equations with different right-hand sides combine linearly : if $y_1$ solves $y' + a y = b_1$ and $y_2$ solves $y' + a y = b_2$, then $\lambda_1 y_1 + \lambda_2 y_2$ solves $y' + a y = \lambda_1 b_1 + \lambda_2 b_2$. The second --- the affine-line structure of the solution set --- says that, knowing a single particular solution $y_{\mathrm{part}}$, every solution is $y_{\mathrm{part}}$ plus a solution of the homogeneous equation. To find a particular solution, the standard technique is variation of constants : replace the constant $\lambda$ in the homogeneous solution $\lambda e^{-A}$ by a function $\lambda(x)$, and pin down $\lambda$ by injecting into the complete equation.

Proposition — Superposition principle at first order

Let $a \in C(I, \mathbb{K})$ and $b_1, b_2 \in C(I, \mathbb{K})$. If $y_1$ solves $y' + a(x) y = b_1(x)$ on $I$ and $y_2$ solves $y' + a(x) y = b_2(x)$ on $I$, then for all $\lambda_1, \lambda_2 \in \mathbb{K}$, the function $\lambda_1 y_1 + \lambda_2 y_2$ solves $$ y'(x) + a(x) \, y(x) \ = \ \lambda_1 b_1(x) + \lambda_2 b_2(x) \qquad \text{on } I. $$

Proof

By linearity of the derivative and of the multiplication by $a$, $$ (\lambda_1 y_1 + \lambda_2 y_2)' + a (\lambda_1 y_1 + \lambda_2 y_2) \ = \ \lambda_1 (y_1' + a y_1) + \lambda_2 (y_2' + a y_2) \ = \ \lambda_1 b_1 + \lambda_2 b_2. $$

Proposition — Affine-line structure of the solution set

Let $a, b \in C(I, \mathbb{K})$ and assume the equation $y' + a(x) y = b(x)$ admits at least one solution $y_{\mathrm{part}} \in C^1(I, \mathbb{K})$. Then $$ S \ = \ y_{\mathrm{part}} + S_0 \ = \ \bigl\{ \, y_{\mathrm{part}} + \lambda \, e^{-A} \;\big|\; \lambda \in \mathbb{K} \, \bigr\}, $$ where $A$ is any primitive of $a$ on $I$. In particular, $S$ is an affine line of direction $S_0 = \mathbb{K} \cdot e^{-A}$.

Proof

Direct corollary of superposition. Let $y \in S$ be any solution of the complete equation ; then $y$ solves $y' + a y = b$ and $y_{\mathrm{part}}$ solves $y' + a y = b$, so by superposition (with $\lambda_1 = 1$, $\lambda_2 = -1$), $y - y_{\mathrm{part}}$ solves $y' + a y = 0$, i.e.\ $y - y_{\mathrm{part}} \in S_0$. Conversely, for any $y_0 \in S_0$, superposition gives $(y_{\mathrm{part}} + y_0)' + a (y_{\mathrm{part}} + y_0) = b + 0 = b$, so $y_{\mathrm{part}} + y_0 \in S$.

Method — Variation of constants

To find a particular solution of $y' + a(x) y = b(x)$ on $I$, in 4 steps :

Solve the homogeneous equation : write a nonzero homogeneous solution $y_h(x) = e^{-A(x)}$, with $A$ a primitive of $a$.
Ansatz : seek $y(x) = \lambda(x) \, e^{-A(x)}$ with $\lambda \in C^1(I, \mathbb{K})$ to be determined.
Inject in the complete equation : compute $y'(x) = \lambda'(x) e^{-A(x)} - \lambda(x) a(x) e^{-A(x)}$, then $y'(x) + a(x) y(x) = \lambda'(x) e^{-A(x)}$ ; setting this equal to $b(x)$ gives $\lambda'(x) = b(x) e^{A(x)}$.
Take a primitive of $b e^A$ to obtain $\lambda$, then assemble $y(x) = \lambda(x) e^{-A(x)}$.

The whole procedure uses two antidifferentiations : one for $A$ (primitive of $a$), one for $\lambda$ (primitive of $b e^A$).

Proposition — Existence of a particular solution

Let $a, b \in C(I, \mathbb{K})$. The equation $y' + a(x) y = b(x)$ admits at least one solution on $I$ : explicitly, if $A$ is a primitive of $a$ on $I$ and $\Lambda$ is a primitive of $b e^A$ on $I$, then $x \mapsto \Lambda(x) \, e^{-A(x)}$ is a solution.

Proof

Both $A$ and $\Lambda$ exist as primitives of continuous functions on an interval (chapter Primitives). Setting $y(x) = \Lambda(x) e^{-A(x)}$, we get $$ \begin{aligned} y'(x) \ &= \ \Lambda'(x) \, e^{-A(x)} - \Lambda(x) \, A'(x) \, e^{-A(x)} && \text{(product + chain rule)} \\ &= \ b(x) \, e^{A(x)} \, e^{-A(x)} - a(x) \, \Lambda(x) \, e^{-A(x)} && \text{($\Lambda' = b e^A$, $A' = a$)} \\ &= \ b(x) - a(x) \, y(x). \end{aligned} $$ Hence $y' + a y = b$ on $I$.

Example — Solving $x y' + y \equal x^2$ on $\mathbb R_+^*$

On $\mathbb R_+^*$, divide by $x$ to put the equation in normal form : $y' + y/x = x$. So $a(x) = 1/x$ and $b(x) = x$. A primitive of $a$ on $\mathbb R_+^*$ is $A(x) = \ln x$, hence $e^{-A(x)} = 1/x$. The homogeneous solutions are $x \mapsto \lambda/x$, $\lambda \in \mathbb R$.
Variation of constants. Ansatz $y(x) = \lambda(x)/x$. Then $y'(x) = \lambda'(x)/x - \lambda(x)/x^2$ and $y'(x) + y(x)/x = \lambda'(x)/x$. Setting this equal to $x$ gives $\lambda'(x) = x^2$, hence $\lambda(x) = x^3/3$ works. A particular solution is $y_{\mathrm{part}}(x) = x^2/3$. The general solution is $$ y(x) \ = \ \dfrac{x^2}{3} + \dfrac{\lambda}{x}, \qquad \lambda \in \mathbb R. $$ Verification : $\bigl( x^2/3 + \lambda/x \bigr)' \cdot x + (x^2/3 + \lambda/x) = 2 x^2/3 - \lambda/x + x^2/3 + \lambda/x = x^2$. $\checkmark$

Example — Constant coefficient with exponential right-hand side

On $\mathbb R$, solve $y' + 2 y = e^x$. Here $a = 2$ constant, $A(x) = 2 x$, $e^{-A(x)} = e^{-2 x}$. Homogeneous solutions $\lambda e^{-2 x}$.
Variation of constants. Ansatz $y(x) = \lambda(x) e^{-2 x}$, then $\lambda'(x) = b(x) e^{A(x)} = e^x \cdot e^{2 x} = e^{3 x}$, so $\lambda(x) = e^{3 x}/3$ and $y_{\mathrm{part}}(x) = e^{3 x}/3 \cdot e^{-2 x} = e^x/3$.
General solution : $y(x) = \lambda e^{-2 x} + e^x/3$, $\lambda \in \mathbb R$. Verification : $(e^x/3)' + 2 (e^x/3) = e^x/3 + 2 e^x/3 = e^x$. $\checkmark$

Skills to practice

Solving complete first-order ODEs by variation of constants

I.3 Cauchy problem (first-order)

Together with the section on Complete equation, superposition, variation of constants, the variation-of-constants computation has actually solved the Cauchy problem : for any $(x_0, y_0) \in I \times \mathbb{K}$, the system $\{y' + a(x) y = b(x), \ y(x_0) = y_0\}$ has exactly one solution on $I$. The closed-form expression is the variation-of-constants formula with the initial condition pinning down the constant. We state it in primitive-first form ; the equivalent definite-integral form is given as a Remark with a forward-pointer to chapter Integration over a segment (chapter 26).

Theorem — Cauchy problem at first order

Let $a, b \in C(I, \mathbb{K})$, $x_0 \in I$, $y_0 \in \mathbb{K}$. Let $A$ be a primitive of $a$ on $I$, and let $B$ be the unique primitive of $b \cdot e^{A}$ on $I$ such that $B(x_0) = 0$. The Cauchy problem $$ \left\{ \begin{aligned} y'(x) + a(x) \, y(x) \ &= \ b(x) \quad \text{on } I, \\ y(x_0) \ &= \ y_0, \end{aligned} \right. $$ has a unique solution on $I$, given by $$ y(x) \ = \ \bigl( y_0 \, e^{A(x_0)} + B(x) \bigr) \, e^{-A(x)}. $$

Proof

By the variation-of-constants method (Complete equation, superposition, variation of constants), every solution of $y' + a y = b$ on $I$ writes $y(x) = \lambda(x) e^{-A(x)}$ where $\lambda$ is a primitive of $b e^A$. Let $B$ be the unique such primitive with $B(x_0) = 0$ (well-defined : two primitives of $b e^A$ differ by a constant, and prescribing the value at $x_0$ pins down that constant). The general primitive of $b e^A$ is then $\lambda(x) = B(x) + c$ for some $c \in \mathbb{K}$, and the corresponding solution reads $$ y(x) \ = \ (B(x) + c) \, e^{-A(x)}. $$ The initial condition $y(x_0) = y_0$ becomes $(B(x_0) + c) e^{-A(x_0)} = c \, e^{-A(x_0)} = y_0$, hence $c = y_0 \, e^{A(x_0)}$. The solution is $$ y(x) \ = \ \bigl( B(x) + y_0 \, e^{A(x_0)} \bigr) \, e^{-A(x)}. $$ Existence : the function above is $C^1$ on $I$ (as a product of $C^1$ functions) and satisfies both the equation (by the variation-of-constants method) and the initial condition (by construction of $c$). Uniqueness : any other solution must also write $\lambda(x) e^{-A(x)}$ with $\lambda$ a primitive of $b e^A$, and the initial condition pins down the constant of integration uniquely.

Definite-integral form (forward-pointer to chapter 26)

In chapter Integration over a segment (chapter 26, semester 2), the primitive $B$ of $b e^A$ vanishing at $x_0$ admits the explicit integral expression $$ B(x) \ = \ \int_{x_0}^{x} b(t) \, e^{A(t)} \, dt, $$ and the unique solution of the Cauchy problem reads $$ y(x) \ = \ y_0 \, e^{A(x_0) - A(x)} + \int_{x_0}^{x} b(t) \, e^{A(t) - A(x)} \, dt. $$ At this stage of the curriculum, this is notation only for the primitive $B$ above ; the rigorous definition of the bounded integral $\int_{x_0}^{x}$ is the subject of chapter 26.

Method — Solving a Cauchy problem at first order

To solve $\{y' + a(x) y = b(x), \ y(x_0) = y_0\}$ on $I$ :

Compute a primitive $A$ of $a$ on $I$ (one antidifferentiation).
Write the homogeneous solutions $\lambda \, e^{-A}$.
Find a particular solution $y_{\mathrm{part}}$ by variation of constants (second antidifferentiation, of $b e^A$).
Write the general solution $y_{\mathrm{part}} + \lambda \, e^{-A}$, then impose $y(x_0) = y_0$ to determine $\lambda$ uniquely.

Example — Cauchy problem with initial condition

On $\mathbb R$, solve $\{y' + y = e^x, \ y(0) = 1\}$. Here $a = 1$ constant, $A(x) = x$, $e^{-A(x)} = e^{-x}$. Homogeneous solutions $\lambda e^{-x}$.
Particular solution : variation of constants gives $\lambda'(x) = e^x \cdot e^x = e^{2 x}$, hence $\lambda(x) = e^{2 x}/2$ and $y_{\mathrm{part}}(x) = e^{2 x}/2 \cdot e^{-x} = e^x/2$.
General solution : $y(x) = \lambda \, e^{-x} + e^x/2$.
Initial condition : $y(0) = \lambda + 1/2 = 1$ gives $\lambda = 1/2$. The unique solution is $$ y(x) \ = \ \dfrac{e^{-x} + e^x}{2} \ = \ \cosh x. $$ Verification : $(\cosh x)' + \cosh x = \sinh x + \cosh x = e^x$ and $\cosh 0 = 1$. $\checkmark$

Skills to practice

Solving Cauchy problems at first order

I.4 Raccord on $a(x) y' + b(x) y \equal f(x)$ --- optional enrichment

The equations in the preceding first-order sections are normalised : the coefficient of $y'$ is $1$. When the actual problem has the form $a(x) y' + b(x) y = f(x)$ and the function $a$ vanishes at isolated points, the raccord (or recollement, the two French terms are synonymous) problem appears : solve on each connected component of $\{a \ne 0\}$ first, then glue the local solutions across the zeros by imposing $C^1$ regularity. The resulting global solution set on the full interval can have smaller or larger dimension than one would guess from a single normalised interval. This subsection is an optional enrichment / exercice classique, not part of the program's core first-order list ; it is included because it is a classical exercise in problem sets and concours preparation, and because it sharpens the « an ODE lives on an interval, not on an arbitrary subset » reflex.

Method — Raccord at a zero of the leading coefficient $a$ --- optional enrichment

To solve $a(x) y' + b(x) y = f(x)$ on an interval $I$ on which $a$ may vanish at isolated points :

Identify the zeros of $a$ on $I$. The connected components of $I \setminus a^{-1}(0)$ are the open intervals on which the normalised equation $y' + (b/a) y = f/a$ makes sense.
Solve $y' + (b/a) y = f/a$ on each connected component (using the homogeneous and variation-of-constants methods above) ; this produces one free constant per component.
For each zero $x_0$ of $a$, require the candidate global function to be of class $C^1$ at $x_0$ : this imposes one or more relations between the per-component constants (limits of the function and of its derivative must agree on both sides of $x_0$).
Verify that the resulting global function actually satisfies the original equation $a(x) y' + b(x) y = f(x)$ at the zeros (often forced by substitution at $x = x_0$).

Example — Solving $x y' + y \equal x^2$ on $\mathbb R$ --- loss of dimension

The coefficient $a(x) = x$ vanishes only at $x = 0$. On each of $\mathbb R_+^*$ and $\mathbb R_-^*$, the normalised equation $y' + y/x = x$ was solved in the variation-of-constants section : the solutions are $y(x) = x^2/3 + \lambda/x$ on $\mathbb R_+^*$ and $y(x) = x^2/3 + \mu/x$ on $\mathbb R_-^*$, with $\lambda, \mu \in \mathbb R$ independent.
Gluing at $x = 0$. For a global $C^1$ function on $\mathbb R$, we need $y(x) \to y(0)$ and $y'(x) \to y'(0)$ as $x \to 0$ from both sides. The term $\lambda/x$ (resp.\ $\mu/x$) is bounded near $0$ only if $\lambda = 0$ (resp.\ $\mu = 0$). Hence the only candidate global solution is $y(x) = x^2/3$, defined and $C^1$ on $\mathbb R$.
Verification at $x = 0$. Substituting $y = x^2/3$ in $x y' + y$ gives $x \cdot (2 x/3) + x^2/3 = x^2$. $\checkmark$ The unique global solution on $\mathbb R$ is $y(x) = x^2/3$.
The local solution sets on $\mathbb R_+^*$ and $\mathbb R_-^*$ are each affine lines of dimension $1$ ; gluing them at $0$ collapses the solution set on $\mathbb R$ to a singleton (dimension $0$). This is a loss of dimension caused by the raccord.

Gain of dimension --- reserved for the exo

The equation $x y' - 2 y = x^3$ on $\mathbb R$ illustrates the opposite phenomenon : the local solutions on $\mathbb R_+^*$ and $\mathbb R_-^*$ are each affine lines (dimension $1$), but the global solution set on $\mathbb R$ is an affine plane (dimension $2$). This example is treated in the exo file.

Skills to practice

Solving raccord (connection) problems

II Second-order linear ODEs with constant coefficients

II.1 Definition$\virgule$ complex lemma$\virgule$ and real homogeneous case

A second-order linear ODE with constant coefficients has the form $y'' + a y' + b y = f(x)$, with $a, b \in \mathbb{K}$ scalars and $f \in C(I, \mathbb{K})$. The characteristic polynomial is $\chi(X) = X^2 + a X + b$ ; its roots drive the structure of the homogeneous solutions. To handle the case of complex conjugate roots cleanly, we first state the complex solution structure as a short bookkeeping lemma over $\mathbb{K} = \mathbb C$, and then read off the real result by taking real parts. For $\mathbb{K} = \mathbb R$, three cases occur depending on the sign of the discriminant $\Delta = a^2 - 4 b$.

Definition — Linear second-order constant-coefficient ODE

Let $a, b \in \mathbb{K}$ scalars and $f \in C(I, \mathbb{K})$. The linear second-order constant-coefficient ODE associated with $(a, b, f)$ is the equation $$ y''(x) + a \, y'(x) + b \, y(x) \ = \ f(x), \qquad x \in I, $$ of unknown $y \in C^2(I, \mathbb{K})$. The associated homogeneous equation is $y'' + a y' + b y = 0$. The characteristic polynomial is $\chi(X) = X^2 + a X + b \in \mathbb{K}[X]$. A solution on $I$ is a function $y \in C^2(I, \mathbb{K})$ satisfying the equation pointwise.

Proposition — Complex homogeneous lemma

Let $a, b \in \mathbb C$. The complex solutions of $y'' + a y' + b y = 0$ on $I$ form a $2$-dimensional $\mathbb C$-vector subspace of $C^2(I, \mathbb C)$. Explicitly, let $r_1, r_2$ denote the (complex) roots of the characteristic polynomial $X^2 + a X + b$ :

if $r_1 \ne r_2$ (two distinct roots), the solutions are $x \mapsto \lambda \, e^{r_1 x} + \mu \, e^{r_2 x}$ with $\lambda, \mu \in \mathbb C$ ;
if $r_1 = r_2 = r$ (double root), the solutions are $x \mapsto (\lambda x + \mu) \, e^{r x}$ with $\lambda, \mu \in \mathbb C$.

Proof

Reduce-order trick. Pick $r$ a root of $\chi(X) = X^2 + a X + b$. We seek the solutions of the homogeneous equation under the form $y(x) = z(x) \, e^{r x}$ with $z \in C^2(I, \mathbb C)$ to be determined. Computing : $$ \begin{aligned} y'(x) \ &= \ (z'(x) + r \, z(x)) \, e^{r x} && \text{(product rule)} \\ y''(x) \ &= \ (z''(x) + 2 r \, z'(x) + r^2 \, z(x)) \, e^{r x} && \text{(differentiating $y'$)}. \end{aligned} $$ Injecting in the homogeneous equation $y'' + a y' + b y = 0$ and factoring $e^{r x}$ (which never vanishes) : $$ z''(x) + (2 r + a) \, z'(x) + \underbrace{(r^2 + a r + b)}_{= \, \chi(r) \, = \, 0} \, z(x) \ = \ 0, $$ which simplifies to $z'' + (2 r + a) z' = 0$, i.e.\ a first-order linear ODE in $u = z'$ : $u' + (2 r + a) u = 0$.

Case $r_1 \ne r_2$. The other root is $r_2 = -a - r$ (since the sum of roots is $-a$), so $2 r + a = r - r_2$. The first-order equation in $u$ is $u' + (r - r_2) u = 0$, with solutions $u(x) = c \, e^{-(r - r_2) x} = c \, e^{(r_2 - r) x}$, $c \in \mathbb C$. Taking a primitive, $z(x) = \dfrac{c}{r_2 - r} e^{(r_2 - r) x} + d$ for $d \in \mathbb C$. Hence $$ y(x) \ = \ z(x) \, e^{r x} \ = \ \dfrac{c}{r_2 - r} e^{r_2 x} + d \, e^{r x}, $$ which (renaming constants) is exactly $\lambda \, e^{r_1 x} + \mu \, e^{r_2 x}$ with $r_1 = r$ (and $\lambda, \mu \in \mathbb C$ arbitrary).
Case $r_1 = r_2 = r$. Then $2 r + a = 0$ (the double root is $r = -a/2$, and indeed $2 r + a = 0$). The first-order equation in $u$ becomes $u' = 0$, hence $u = c$ constant. Taking a primitive, $z(x) = c x + d$ for $c, d \in \mathbb C$. Hence $$ y(x) \ = \ z(x) \, e^{r x} \ = \ (c x + d) \, e^{r x}, $$ which (renaming constants) is $(\lambda x + \mu) e^{r x}$.

In both cases the solution set has dimension $2$ over $\mathbb C$.

Theorem — Real homogeneous equation $y'' + a y' + b y \equal 0$ for $\mathbb{K} \equal \mathbb R$

Let $a, b \in \mathbb R$. The real solutions of $y'' + a y' + b y = 0$ on $I$ form a $2$-dimensional $\mathbb R$-vector subspace of $C^2(I, \mathbb R)$ described by the discriminant $\Delta = a^2 - 4 b$ of $X^2 + a X + b$ :

$\Delta > 0$ (two distinct real roots $r \ne r'$) : solutions $x \mapsto \lambda \, e^{r x} + \mu \, e^{r' x}$, $\lambda, \mu \in \mathbb R$ ;
$\Delta = 0$ (double real root $r = -a/2$) : solutions $x \mapsto (\lambda x + \mu) \, e^{r x}$, $\lambda, \mu \in \mathbb R$ ;
$\Delta < 0$ (complex conjugate roots $r \pm i \omega$ with $r = -a/2$ and $\omega = \sqrt{-\Delta}/2 > 0$) : solutions $x \mapsto e^{r x} \bigl( \lambda \cos(\omega x) + \mu \sin(\omega x) \bigr)$, $\lambda, \mu \in \mathbb R$.

Proof

The student must know the table by heart ; the proof is for understanding. The $\Delta > 0$ and $\Delta = 0$ cases follow immediately from the complex homogeneous lemma above, since the roots are real and the formulas $\lambda e^{r_1 x} + \mu e^{r_2 x}$ and $(\lambda x + \mu) e^{r x}$ are real for $\lambda, \mu \in \mathbb R$.
For the $\Delta < 0$ case, the roots are complex conjugate, $r_1 = r + i \omega$ and $r_2 = r - i \omega$ with $r = -a/2 \in \mathbb R$ and $\omega = \sqrt{-\Delta}/2 > 0$. By the complex lemma, the complex solutions are $y(x) = \alpha \, e^{(r + i \omega) x} + \beta \, e^{(r - i \omega) x}$ with $\alpha, \beta \in \mathbb C$. Factor $e^{r x}$ : $$ y(x) \ = \ e^{r x} \bigl( \alpha \, e^{i \omega x} + \beta \, e^{- i \omega x} \bigr). $$ We seek the condition for $y$ to be real-valued on $I$, i.e.\ $y = \overline{y}$ on $I$. Conjugating term-by-term in the expression above, $$ \overline{y}(x) \ = \ \overline{\alpha} \, e^{(r - i \omega) x} + \overline{\beta} \, e^{(r + i \omega) x}. $$ The equality $y = \overline{y}$ on $I$ reads $$ \alpha \, e^{(r + i \omega) x} + \beta \, e^{(r - i \omega) x} \ = \ \overline{\beta} \, e^{(r + i \omega) x} + \overline{\alpha} \, e^{(r - i \omega) x} \quad \text{on } I. $$ The two functions $e^{(r + i \omega) x}$ and $e^{(r - i \omega) x}$ are $\mathbb C$-linearly independent on $I$ : indeed, if they were dependent, their quotient $e^{2 i \omega x}$ would be constant on $I$, which is impossible since its derivative $2 i \omega \, e^{2 i \omega x}$ is non-zero. Identifying coefficients in the equality above therefore forces $\alpha = \overline{\beta}$, i.e.\ $\beta = \overline{\alpha}$. Conversely, if $\beta = \overline{\alpha}$ then $$ \alpha \, e^{i \omega x} + \overline{\alpha} \, e^{- i \omega x} \ = \ 2 \operatorname{Re}(\alpha \, e^{i \omega x}) \ = \ 2 \operatorname{Re}(\alpha) \cos(\omega x) - 2 \operatorname{Im}(\alpha) \sin(\omega x), $$ so setting $\lambda = 2 \operatorname{Re}(\alpha)$ and $\mu = - 2 \operatorname{Im}(\alpha)$, we get $y(x) = e^{r x} (\lambda \cos(\omega x) + \mu \sin(\omega x))$. Both $\lambda, \mu \in \mathbb R$ can be chosen arbitrarily by appropriate choice of $\alpha \in \mathbb C$, so the real solution space has dimension $2$ over $\mathbb R$ with basis $(x \mapsto e^{r x} \cos(\omega x), \ x \mapsto e^{r x} \sin(\omega x))$.

Method — Solving $y'' + a y' + b y \equal 0$ over $\mathbb R$

In 3 steps :

Compute the discriminant $\Delta = a^2 - 4 b$ of $X^2 + a X + b$.
Factor the characteristic polynomial : two distinct real roots if $\Delta > 0$, a double real root if $\Delta = 0$, two complex conjugate roots $r \pm i \omega$ if $\Delta < 0$.
Write the general solution per the table above ($\Delta > 0$ : exponentials ; $\Delta = 0$ : polynomial times exponential ; $\Delta < 0$ : exponential times $\cos / \sin$).

Example — Two distinct real roots ($\Delta > 0$)

Solve $y'' - 3 y' + 2 y = 0$ on $\mathbb R$. Characteristic polynomial $X^2 - 3 X + 2 = (X - 1)(X - 2)$, so $\Delta = 9 - 8 = 1 > 0$ with roots $r = 1$ and $r' = 2$. General solution : $$ y(x) \ = \ \lambda \, e^x + \mu \, e^{2 x}, \qquad \lambda, \mu \in \mathbb R. $$

Example — Double real root ($\Delta \equal 0$)

Solve $y'' - 2 y' + y = 0$ on $\mathbb R$. Characteristic polynomial $X^2 - 2 X + 1 = (X - 1)^2$, double root $r = 1$ ($\Delta = 0$). General solution : $$ y(x) \ = \ (\lambda x + \mu) \, e^x, \qquad \lambda, \mu \in \mathbb R. $$ This is the critically damped regime in physics (e.g.\ for the linearised harmonic oscillator at the boundary between under- and over-damped behaviour).

Example — Complex conjugate roots ($\Delta < 0$)

Pure imaginary roots (the harmonic oscillator). Solve $y'' + 4 y = 0$ on $\mathbb R$. Characteristic polynomial $X^2 + 4$, $\Delta = -16 < 0$, roots $\pm 2 i$ (so $r = 0$ and $\omega = 2$). General solution : $$ y(x) \ = \ \lambda \cos(2 x) + \mu \sin(2 x), \qquad \lambda, \mu \in \mathbb R. $$ Roots with non-zero real part (damped oscillator). Solve $y'' + 2 y' + 10 y = 0$ on $\mathbb R$. Characteristic polynomial $X^2 + 2 X + 10 = (X + 1)^2 + 9$, $\Delta = 4 - 40 = -36 < 0$, roots $-1 \pm 3 i$ (so $r = -1$ and $\omega = 3$). General solution : $$ y(x) \ = \ e^{-x} \bigl( \lambda \cos(3 x) + \mu \sin(3 x) \bigr), \qquad \lambda, \mu \in \mathbb R. $$ The $e^{-x}$ factor encodes the exponential damping ; the $\cos(3 x), \sin(3 x)$ part the oscillation at angular frequency $3$.

Skills to practice

Solving homogeneous second-order ODEs

II.2 Complete equation$\virgule$ superposition$\virgule$ Cauchy (admitted)

The complete equation $y'' + a y' + b y = f(x)$ has solution set $S = y_{\mathrm{part}} + S_0$, an affine plane. The superposition principle holds as it did for the first-order complete equation : if $y_k$ solves $y'' + a y' + b y = f_k$ for $k = 1, 2$, then $\lambda_1 y_1 + \lambda_2 y_2$ solves the equation with right-hand side $\lambda_1 f_1 + \lambda_2 f_2$. The Cauchy problem at second order requires two initial conditions : $y(x_0) = y_0$ and $y'(x_0) = y_0'$. Its existence-and-uniqueness theorem is stated here and admitted ; the proof is out of program.

Proposition — Superposition principle at second order

Let $a, b \in \mathbb{K}$ and $f_1, f_2 \in C(I, \mathbb{K})$. If $y_1$ solves $y'' + a y' + b y = f_1(x)$ on $I$ and $y_2$ solves $y'' + a y' + b y = f_2(x)$ on $I$, then for all $\lambda_1, \lambda_2 \in \mathbb{K}$, the function $\lambda_1 y_1 + \lambda_2 y_2$ solves $$ y''(x) + a \, y'(x) + b \, y(x) \ = \ \lambda_1 f_1(x) + \lambda_2 f_2(x) \qquad \text{on } I. $$

Proof

By linearity of differentiation and of the multiplication by $a, b$, $$ (\lambda_1 y_1 + \lambda_2 y_2)'' + a (\lambda_1 y_1 + \lambda_2 y_2)' + b (\lambda_1 y_1 + \lambda_2 y_2) \ = \ \lambda_1 (y_1'' + a y_1' + b y_1) + \lambda_2 (y_2'' + a y_2' + b y_2) \ = \ \lambda_1 f_1 + \lambda_2 f_2. $$

Proposition — Affine-plane structure of the solution set

Let $a, b \in \mathbb{K}$, $f \in C(I, \mathbb{K})$, and assume the equation $y'' + a y' + b y = f(x)$ admits at least one solution $y_{\mathrm{part}} \in C^2(I, \mathbb{K})$. Then $$ S \ = \ y_{\mathrm{part}} + S_0, $$ where $S_0$ is the $2$-dimensional space of homogeneous solutions. In particular, $S$ is an affine plane in $C^2(I, \mathbb{K})$.

Proof

Direct corollary of superposition, mirroring the first-order complete-equation argument. Let $y \in S$ be any solution of the complete equation ; then $y$ and $y_{\mathrm{part}}$ both solve the same complete equation, so by superposition (with $\lambda_1 = 1$, $\lambda_2 = -1$), $y - y_{\mathrm{part}}$ solves the homogeneous equation, i.e.\ $y - y_{\mathrm{part}} \in S_0$. Conversely, for any $y_0 \in S_0$, superposition gives $(y_{\mathrm{part}} + y_0)'' + a (y_{\mathrm{part}} + y_0)' + b (y_{\mathrm{part}} + y_0) = f + 0 = f$, so $y_{\mathrm{part}} + y_0 \in S$.

Theorem — Cauchy problem at second order

Let $a, b \in \mathbb{K}$, $f \in C(I, \mathbb{K})$, $x_0 \in I$, and $(y_0, y_0') \in \mathbb{K}^2$. The Cauchy problem $$ \left\{ \begin{aligned} y''(x) + a \, y'(x) + b \, y(x) \ &= \ f(x) \quad \text{on } I, \\ y(x_0) \ &= \ y_0, \\ y'(x_0) \ &= \ y_0', \end{aligned} \right. $$ has a unique solution on $I$. In particular, taking arbitrary initial data gives the existence of at least one particular solution of the complete equation, which retroactively validates the affine-plane structure stated above.

Proof

Admitted at this level; see the Remark below for the proof status.

On the proof status

This theorem is admitted : the proof is out of program (programme 2021 p. 12, « la démonstration de ce résultat est hors programme »). A constructive proof reduces the second-order problem to a first-order Cauchy problem via the substitution $y = z e^{r x}$ with $r$ a root of the characteristic polynomial, then applies the first-order Cauchy theorem proven above.

Method — Solving a Cauchy problem at second order

To solve $\{y'' + a y' + b y = f(x), \ y(x_0) = y_0, \ y'(x_0) = y_0'\}$ on $I$ :

Solve the homogeneous equation $y'' + a y' + b y = 0$ (real homogeneous case) : $2$-parameter family $S_0$.
Find a particular solution $y_{\mathrm{part}}$ of the complete equation using the polynomial-exponential ansatz.
Write the general solution $y(x) = y_{\mathrm{part}}(x) + y_h(x)$ with $y_h \in S_0$ (two free parameters).
Impose the two initial conditions $y(x_0) = y_0$ and $y'(x_0) = y_0'$ : a linear system in the two free parameters, with a unique solution by the theorem.

Example — Cauchy problem with two conditions

On $\mathbb R$, solve $\{y'' - y = 2 \cos x + e^x, \ y(0) = -1, \ y'(0) = 0\}$.
Homogeneous part. Characteristic polynomial $X^2 - 1 = (X - 1)(X + 1)$, two distinct real roots $\pm 1$, $\Delta = 4 > 0$. Homogeneous solutions $\lambda e^x + \mu e^{-x}$.
Particular solution by superposition. Split the right-hand side into $f_1(x) = 2 \cos x$ and $f_2(x) = e^x$, then sum particular solutions.

For $y'' - y = 2 \cos x$, use the complex method from the practical-recipes section : seek a complex solution of $y'' - y = 2 e^{i x}$ as $y_p(x) = \alpha e^{i x}$. Computing $y_p'' - y_p = (-1 - 1) \alpha e^{i x} = - 2 \alpha e^{i x} = 2 e^{i x}$ gives $\alpha = -1$, so $y_p(x) = - e^{i x}$. The real solution of $y'' - y = 2 \cos x$ is $\operatorname{Re}(- e^{i x}) = -\cos x$.
For $y'' - y = e^x$, here $P(x) = 1$ has degree $0$, and $u = 1$ is a simple root of $X^2 - 1$ ($m = 1$). The ansatz is $y_p(x) = x \, Q(x) \, e^x$ with $Q$ of degree at most $0$, i.e.\ $Q(x) = a$ constant : $y_p(x) = a x e^x$. Computing $y_p''(x) = a (x e^x + 2 e^x)$, so $y_p'' - y_p = 2 a e^x = e^x$ gives $a = 1/2$. Hence $y_p(x) = x e^x / 2$.

By superposition, a particular solution of the complete equation is $y_{\mathrm{part}}(x) = - \cos x + x e^x / 2$.
General solution. $y(x) = \lambda e^x + \mu e^{-x} + x e^x/2 - \cos x$.
Initial conditions. Compute $y'(x) = \lambda e^x - \mu e^{-x} + e^x/2 + x e^x/2 + \sin x$. Then $$ \begin{aligned} y(0) \ &= \ \lambda + \mu - 1 \ = \ - 1 && \Longrightarrow \quad \lambda + \mu \ = \ 0, \\ y'(0) \ &= \ \lambda - \mu + 1/2 \ = \ 0 && \Longrightarrow \quad \lambda - \mu \ = \ -1/2, \end{aligned} $$ so $\lambda = -1/4$, $\mu = 1/4$. The unique solution is $$ y(x) \ = \ \dfrac{(2 x - 1) \, e^x + e^{-x}}{4} - \cos x. $$

Skills to practice

Solving second-order Cauchy problems

II.3 Practical recipes$\virgule$ polynomial-exponential ansatz

The program lists three categories of right-hand side the student must master : polynomial, $\alpha \, e^{\lambda x}$ with $(\alpha, \lambda) \in \mathbb C^2$, and $\beta \cos(\omega x) / \beta \sin(\omega x)$ with $(\beta, \omega) \in \mathbb R^2$. All three are subsumed by the unified ansatz, stated over $\mathbb C$ first : for a (possibly complexified) right-hand side of the form $P(x) \, e^{u x}$ with $P \in \mathbb C[X]$ and $u \in \mathbb C$, seek a complex particular solution of the form $x^m \, Q(x) \, e^{u x}$ where $Q \in \mathbb C[X]$ has degree at most $\deg P$, and $m \in \{0, 1, 2\}$ is the multiplicity of $u$ as a root of the characteristic polynomial $X^2 + a X + b$. For a real equation with a trigonometric right-hand side, complexify ($\beta \cos(\omega x) = \operatorname{Re}(\beta e^{i \omega x})$), solve the complexified equation by the same ansatz, then take real (resp.\ imaginary) parts at the end. This is the same $\operatorname{Re}(e^{(a + i b) x})$ reflex from the chapter Primitives.

Method — Polynomial-exponential ansatz over $\mathbb C$

For the complexified equation $y'' + a y' + b y = P(x) \, e^{u x}$ with $P \in \mathbb C[X]$ non-zero and $u \in \mathbb C$ :

Compute the multiplicity $m \in \{0, 1, 2\}$ of $u$ as a root of the characteristic polynomial $X^2 + a X + b$ ($m = 0$ if $u$ is not a root, $m = 1$ if simple root, $m = 2$ if double root).
Seek a complex particular solution of the form $$ y_p(x) \ = \ x^m \, Q(x) \, e^{u x}, $$ where $Q \in \mathbb C[X]$ has degree at most $\deg P$.
In practice, take $Q$ as a generic polynomial of degree $\deg P$, inject the ansatz in the equation, expand, and identify coefficients of like powers of $x$ to determine $Q$.

For a real equation with a trigonometric right-hand side of the form $\beta \cos(\omega x)$ or $\beta \sin(\omega x)$ : complexify via $\beta \cos(\omega x) = \operatorname{Re}(\beta e^{i \omega x})$ (resp.\ $\operatorname{Im}$ for $\sin$), apply the above ansatz to the complexified problem with $u = i \omega$, then take the real (resp.\ imaginary) part of $y_p$ at the end to obtain a real particular solution.

Example — Polynomial right-hand side ($m \equal 0$)

On $\mathbb R$, solve $y'' - 4 y = x^2 + 1$. The right-hand side $x^2 + 1$ has the form $P(x) \, e^{0 \cdot x}$ with $P(x) = x^2 + 1$ and $u = 0$. Since $0$ is not a root of $X^2 - 4 = (X - 2)(X + 2)$, $m = 0$. The ansatz $y_p(x) = x^0 \, Q(x) \, e^{0 \cdot x} = Q(x)$ with $Q$ of degree at most $2$ : $y_p(x) = a x^2 + b x + c$ with $a, b, c \in \mathbb R$. Computing $y_p''(x) = 2 a$, so $$ y_p'' - 4 y_p \ = \ 2 a - 4 (a x^2 + b x + c) \ = \ - 4 a \, x^2 - 4 b \, x + (2 a - 4 c). $$ Identification with $x^2 + 1$ gives $- 4 a = 1$ (so $a = - 1/4$), $- 4 b = 0$ ($b = 0$), and $2 a - 4 c = 1$ (so $c = (2 a - 1)/4 = - 3/8$). A particular solution is $y_p(x) = - x^2/4 - 3/8$. General solution : $y(x) = \lambda e^{2 x} + \mu e^{- 2 x} - x^2/4 - 3/8$.

Example — Exponential right-hand side with resonance ($m \equal 1$)

On $\mathbb R$, solve $y'' - 3 y' + 2 y = (x + 2) e^x$. The characteristic polynomial is $X^2 - 3 X + 2 = (X - 1)(X - 2)$, with simple roots $1$ and $2$. The right-hand side is $P(x) e^{u x}$ with $P(x) = x + 2$ (degree $1$) and $u = 1$, which is a simple root of the characteristic polynomial, so $m = 1$.
By the ansatz, we seek $y_p(x) = x \, Q(x) \, e^x$ with $Q$ of degree at most $1$, i.e.\ $Q(x) = a x + b$. So $$ y_p(x) \ = \ x (a x + b) \, e^x \ = \ (a x^2 + b x) \, e^x. $$ Compute : $$ \begin{aligned} y_p'(x) \ &= \ (2 a x + b) e^x + (a x^2 + b x) e^x \ = \ (a x^2 + (2 a + b) x + b) e^x, \\ y_p''(x) \ &= \ (2 a x + 2 a + b) e^x + (a x^2 + (2 a + b) x + b) e^x \ = \ (a x^2 + (4 a + b) x + (2 a + 2 b)) e^x. \end{aligned} $$ Then $$ y_p'' - 3 y_p' + 2 y_p \ = \ e^x \bigl( a x^2 (1 - 3 + 2) + x (4 a + b - 3 (2 a + b) + 2 b) + (2 a + 2 b - 3 b) \bigr) \ = \ e^x \bigl( - 2 a \, x + (2 a - b) \bigr). $$ Identification with $(x + 2) e^x$ gives $- 2 a = 1$ ($a = -1/2$) and $2 a - b = 2$ ($b = 2 a - 2 = -3$). A particular solution is $y_p(x) = (- x^2/2 - 3 x) e^x = -\dfrac{x (x + 6)}{2} \, e^x$. General solution : $y(x) = \lambda e^x + \mu e^{2 x} -\dfrac{x (x + 6)}{2} \, e^x$.

Example — Trigonometric right-hand side with resonance ($m \equal 1$) via the complex method

On $\mathbb R$, solve $y'' + 4 y = 3 \sin(2 x)$. The characteristic polynomial is $X^2 + 4 = (X - 2 i)(X + 2 i)$, with simple roots $\pm 2 i$.
Complexification. Write $3 \sin(2 x) = \operatorname{Im}(3 \, e^{2 i x})$. Solve the complexified equation $y'' + 4 y = 3 \, e^{2 i x}$. Here $P(x) = 3$ has degree $0$, and $u = 2 i$ is a simple root of the characteristic polynomial, so $m = 1$. The ansatz is $y_p(x) = x \, Q(x) \, e^{2 i x}$ with $Q$ of degree at most $0$, i.e.\ $Q(x) = a$ a complex constant. So $$ y_p(x) \ = \ a \, x \, e^{2 i x}, \qquad a \in \mathbb C. $$ Compute : $$ \begin{aligned} y_p'(x) \ &= \ a \, e^{2 i x} + 2 i a x \, e^{2 i x} \ = \ a (1 + 2 i x) \, e^{2 i x}, \\ y_p''(x) \ &= \ 2 i a \, e^{2 i x} + a (1 + 2 i x) \cdot 2 i \, e^{2 i x} \ = \ a \, e^{2 i x} (2 i + 2 i (1 + 2 i x)) \ = \ a \, e^{2 i x} (4 i - 4 x). \end{aligned} $$ Then $$ y_p'' + 4 y_p \ = \ a \, e^{2 i x} (4 i - 4 x + 4 x) \ = \ 4 i a \, e^{2 i x}. $$ Identification with $3 \, e^{2 i x}$ gives $4 i a = 3$, hence $a = 3 / (4 i) = - 3 i/4$. So a complex particular solution is $$ y_p(x) \ = \ - \dfrac{3 i}{4} \, x \, e^{2 i x} \ = \ - \dfrac{3 i}{4} \, x \, (\cos(2 x) + i \sin(2 x)) \ = \ \dfrac{3 x}{4} \, \sin(2 x) - \dfrac{3 i x}{4} \, \cos(2 x). $$ Taking the imaginary part : $\operatorname{Im}(y_p)(x) = - \dfrac{3 x}{4} \, \cos(2 x)$. This is a real particular solution of $y'' + 4 y = 3 \sin(2 x)$. General real solution : $$ y(x) \ = \ \lambda \cos(2 x) + \mu \sin(2 x) - \dfrac{3 x}{4} \, \cos(2 x), \qquad \lambda, \mu \in \mathbb R. $$ The $x \cdot \cos(2 x)$ factor reveals the resonance : the right-hand side oscillates at the same angular frequency $\omega = 2$ as the homogeneous solutions, producing an unbounded particular solution. This is the same phenomenon as the $x e^x$ ansatz in the previous example (multiplicity $m = 1$ in both cases).

Cross-link to chapter Primitives

The complex method here is the same $\operatorname{Re}(e^{(a + i b) x})$ reflex used in the chapter Primitives to compute primitives of $e^{a x} \cos(b x)$ and $e^{a x} \sin(b x)$. The student should recognize the pattern : whenever a real ODE has a $\cos / \sin$ right-hand side, complexify to $e^{i \omega x}$, solve in $\mathbb C$, then take real or imaginary parts at the end.

Skills to practice

Finding particular solutions by polynomial-exponential ansatz