CommeUnJeu · L2 MP

Linear differential equations

⌚ ~67 min ▢ 8 blocks ✓ 13 exercises ➣ Prerequisites : Linear differential equations, Vector-valued functions of a real variable

MPSI solved two linear differential equations: the scalar order-one equation $y' = a(t)\,y + b(t)$, by an explicit primitive-based formula; and the constant-coefficient $y'' + a\,y' + b\,y = c(t)$, by combining the characteristic equation (for the homogeneous part) with a polynomial-exponential ansatz that handles the standard right-hand sides. This chapter steps up to the general linear differential equation $x' = a(t)\cdot x + b(t)$ where the unknown $x$ is a vector-valued function with values in a finite-dimensional normed space $E$.
The headline of \S1 is the linear Cauchy theorem: through each initial condition $(t_0, x_0)$ passes exactly one solution, and that solution is defined on the whole interval $I$. From it follows the structure of the solution set --- an affine space whose direction has dimension $\dim E$. The chapter then turns, in \S2, to the case the program singles out: the scalar equation of order two with possibly non-constant coefficients --- its solution plane, the wronskian that detects a basis of solutions, and the method of variation of constants that produces a particular solution.
Standing notation. Throughout, $\mathbb{K}$ denotes $\mathbb{R}$ or $\mathbb{C}$, $I$ is an interval of $\mathbb{R}$ with non-empty interior (at an endpoint of $I$, « derivable » means the appropriate one-sided derivative, the standard convention of Vector-valued functions of a real variable), and $E$ is a finite-dimensional $\mathbb{K}$-normed space. The space of $\mathbb{K}$-linear maps $\mathcal{L}(E)$ is itself finite-dimensional, and any two norms on it are equivalent (recalled from Compactness, connectedness, finite dimension), so topological notions on $\mathcal{L}(E)$ are unambiguous. $\mathcal{C}^k(I,E)$ denotes the space of $k$-times continuously differentiable maps $I \to E$. For brevity, when $u \in \mathcal{L}(E)$ and $e \in E$ we write $u \cdot e$ for the evaluation $u(e)$. An equation is written $(E)$ and its homogeneous companion $(E_0)$.

I The linear differential equation $x' \equal a(t)\cdot x + b(t)$

I.1 The equation and the Cauchy problem

The MPSI scalar equation $y' = a(t)\,y + b(t)$ involved a scalar coefficient $a(t) \in \mathbb{K}$ multiplying the scalar unknown $y$. In a finite-dimensional space $E$, the coefficient becomes a linear map $a(t) \in \mathcal{L}(E)$ acting on the vector unknown $x \in E$. The right-hand side $b(t)$ is itself a vector. The equation then reads $x' = a(t) \cdot x + b(t)$, where the dot denotes the evaluation of an endomorphism on a vector. Everything else --- the homogeneous case, the initial-condition Cauchy problem, the equivalent integral formulation --- transcribes the MPSI material word for word, with vectors in place of scalars.

Definition — Linear differential equation

Let $I$ be an interval of $\mathbb{R}$ with non-empty interior, and $E$ a finite-dimensional $\mathbb{K}$-normed space. Let $a \in \mathcal{C}(I, \mathcal{L}(E))$ and $b \in \mathcal{C}(I, E)$. The linear differential equation associated with $a$ and $b$ is $$ (E)\colon \quad x' = a(t) \cdot x + b(t). $$ A solution of $(E)$ on $I$ is a map $\varphi \colon I \to E$ of class $\mathcal{C}^1$ satisfying $\varphi'(t) = a(t) \cdot \varphi(t) + b(t)$ for every $t \in I$. When $b = 0$, the equation is the homogeneous companion $$ (E_0)\colon \quad x' = a(t) \cdot x. $$

We adopt the regularity $\mathcal{C}^1$ from the start, following the prépa tradition. The choice is consistent with the structure of the equation: were $\varphi$ only required derivable, the right-hand side $a(t) \cdot \varphi(t) + b(t)$ would still be continuous --- $a$ and $b$ are continuous by hypothesis, the evaluation $(u, e) \mapsto u \cdot e$ is continuous on $\mathcal{L}(E) \times E$ (bilinear on a finite-dimensional space, recalled from Limits and continuity in a normed space), and $\varphi$ derivable is in particular continuous. So $\varphi' = a \cdot \varphi + b$ is continuous, and $\varphi$ is automatically of class $\mathcal{C}^1$.

Example — The scalar case recovered

For $E = \mathbb{K}$, an element of $\mathcal{L}(\mathbb{K}) = \mathbb{K}$ is just a scalar, and the equation $x' = a(t) \cdot x + b(t)$ becomes the scalar order-one equation $y' = a(t)\,y + b(t)$ of MPSI, with continuous $a, b \colon I \to \mathbb{K}$. The MPSI explicit formula --- recalled here for use later in the chapter --- reads $y(t) = e^{A(t)}\bigl(y_0 + \int_{t_0}^{t} e^{-A(s)} b(s)\,ds\bigr)$ for any primitive $A$ of $a$, giving the unique solution with $y(t_0) = y_0$.

Example — A planar homogeneous system

On $E = \mathbb{R}^2$, consider the planar homogeneous system $$ x' = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \cdot x. $$ The map $a \colon \mathbb{R} \to \mathcal{L}(\mathbb{R}^2)$ is the constant endomorphism whose matrix in the canonical basis is the matrix above. One checks by direct differentiation that $\varphi_1 \colon t \mapsto (\cos t, -\sin t)$ and $\varphi_2 \colon t \mapsto (\sin t, \cos t)$ are solutions, and one anticipates that the general solution is the linear combination $\lambda_1 \varphi_1 + \lambda_2 \varphi_2$ --- the dimension theorem of \S 1.3 will confirm this in full generality.

Definition — Cauchy problem

Let $(t_0, x_0) \in I \times E$. The Cauchy problem associated with $(E)$ and the data $(t_0, x_0)$ is the differential equation $(E)$ together with the initial condition $x(t_0) = x_0$: $$ \begin{cases} x' = a(t) \cdot x + b(t), \\ x(t_0) = x_0. \end{cases} $$ A solution of the Cauchy problem is a solution $\varphi$ of $(E)$ on $I$ satisfying $\varphi(t_0) = x_0$.

Example — A planar Cauchy problem

Continuing the previous Example, the Cauchy problem $$ \begin{cases} x' = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \cdot x, \\ x(0) = (1, 0) \end{cases} $$ admits $\varphi \colon t \mapsto (\cos t, -\sin t)$ as a solution on $\mathbb{R}$ (differentiate: $\varphi'(t) = (-\sin t, -\cos t)$; and $a \cdot \varphi(t) = (-\sin t, -\cos t)$ also, since the matrix sends $(\cos t, -\sin t)$ to $(-\sin t, -\cos t)$). At $t = 0$, $\varphi(0) = (1, 0)$. Uniqueness on all of $\mathbb{R}$ will be granted by the linear Cauchy theorem of \S 1.2.

Proposition — Integral formulation of the Cauchy problem

Let $(t_0, x_0) \in I \times E$ and $\varphi \colon I \to E$. The following are equivalent:

[(i)] $\varphi$ is a solution of the Cauchy problem on $I$;
[(ii)] $\varphi$ is continuous on $I$ and satisfies, for every $t \in I$, $$ \varphi(t) = x_0 + \int_{t_0}^{t} \bigl(\,a(s) \cdot \varphi(s) + b(s)\,\bigr) \,\mathrm{d}s. $$

Proof

(i) $\Rightarrow$ (ii). If $\varphi$ is a $\mathcal{C}^1$ solution of the Cauchy problem, then $\varphi$ is in particular continuous. The fundamental theorem of calculus (recalled from Vector-valued functions of a real variable) gives, for every $t \in I$, $$ \varphi(t) - \varphi(t_0) = \int_{t_0}^{t} \varphi'(s) \,\mathrm{d}s = \int_{t_0}^{t} \bigl(\,a(s) \cdot \varphi(s) + b(s)\,\bigr) \,\mathrm{d}s, $$ and $\varphi(t_0) = x_0$, hence the integral identity.
(ii) $\Rightarrow$ (i). Conversely, suppose $\varphi$ is continuous on $I$ and satisfies the integral identity. The integrand $s \mapsto a(s) \cdot \varphi(s) + b(s)$ is continuous on $I$ (composition and sum of continuous maps), so the integral $t \mapsto \int_{t_0}^{t}(a \cdot \varphi + b)$ is of class $\mathcal{C}^1$ on $I$ (fundamental theorem of calculus). Hence $\varphi$ is $\mathcal{C}^1$, with $\varphi'(t) = a(t) \cdot \varphi(t) + b(t)$ for every $t \in I$. Evaluating the integral identity at $t = t_0$ gives $\varphi(t_0) = x_0$. So $\varphi$ solves the Cauchy problem.

Method — Set up a linear differential equation

Faced with a vector-valued unknown $x(t)$ satisfying a differential relation:

identify the working interval $I$ on which the relation is to hold;
read off the linear part $a(t) \in \mathcal{L}(E)$ acting on $x$ --- check that $a$ is continuous on $I$;
read off the right-hand side $b(t) \in E$ --- check that $b$ is continuous on $I$;
write the equation in the standard form $x' = a(t) \cdot x + b(t)$, and the homogeneous companion $(E_0)\colon x' = a(t) \cdot x$.

A Cauchy problem requires in addition a data $(t_0, x_0) \in I \times E$ for the initial condition $x(t_0) = x_0$.

Skills to practice

Setting up a linear differential equation

I.2 The linear Cauchy theorem

The headline existence-and-uniqueness theorem. Geometrically: through each initial condition $(t_0, x_0) \in I \times E$ passes exactly one solution, and that solution is defined on the whole interval $I$ --- no maximal-interval issue, no possibility of « blow-up ». The linear structure of $(E)$ is what makes this so clean.

Theorem — Linear Cauchy theorem

Let $a \in \mathcal{C}(I, \mathcal{L}(E))$ and $b \in \mathcal{C}(I, E)$. For every $(t_0, x_0) \in I \times E$, the Cauchy problem $$ \begin{cases} x' = a(t) \cdot x + b(t), \\ x(t_0) = x_0 \end{cases} $$ admits one and only one solution $\varphi$, and $\varphi$ is defined on the whole interval $I$.

The proof is non-exigible

The official program lists « Cauchy linéaire » among the results whose proof is non exigible. The theorem is therefore stated and admitted in this chapter. The argument that establishes it rests on analytic tools that are not part of the program --- neither the technique nor any auxiliary lemma is developed here.

Example — Two solutions never cross

Two distinct solutions $\varphi_1, \varphi_2$ of $(E)$ on $I$ satisfy $\varphi_1(t) \ne \varphi_2(t)$ for every $t \in I$: if they agreed at some point $t_0$, they would both solve the same Cauchy problem with data $(t_0, \varphi_1(t_0))$, and uniqueness in the linear Cauchy theorem would force $\varphi_1 = \varphi_2$ on $I$. Geometrically, the « integral curves » $\{(t, \varphi(t)) : t \in I\}$ of distinct solutions are disjoint --- they foliate the strip $I \times E$.

Example — A non-zero homogeneous solution vanishes nowhere

Let $\varphi$ be a solution of $(E_0)\colon x' = a(t) \cdot x$ on $I$. If $\varphi$ vanishes at one point $t_0 \in I$, then $\varphi$ solves the same Cauchy problem as the zero function (data $(t_0, 0)$). By uniqueness, $\varphi$ is identically zero on $I$. Contrapositive: a solution of $(E_0)$ that does not vanish at one point does not vanish anywhere on $I$.

Method — Use uniqueness without solving

To identify a candidate function $\tilde\varphi$ (built from a known solution $\varphi$ by some construction --- a scalar multiple $\lambda \varphi$, a translate $t \mapsto \varphi(t+s)$, a reflection $t \mapsto \varphi(-t)$, \ldots) with $\varphi$ itself, verify three things in order:

[(1)] $\tilde\varphi$ is defined on the same interval $I$ as $\varphi$ --- translation or reflection may require an appropriate symmetry of $I$ or restriction to a subinterval;
[(2)] $\tilde\varphi$ satisfies the same differential equation as $\varphi$ on $I$ --- which requires a matching invariance of the coefficients $a, b$ (for instance, for a translate by $s$, the coefficients must themselves be $s$-periodic);
[(3)] at some $t_0 \in I$, $\tilde\varphi(t_0) = \varphi(t_0)$.

The linear Cauchy theorem then forces $\tilde\varphi = \varphi$ on all of $I$. All three conditions are needed --- the equation alone is not enough.

Skills to practice

Exploiting the linear Cauchy theorem

I.3 Structure of the set of solutions

The set of solutions of $(E)$ has a clean algebraic shape: an affine subspace of $\mathcal{C}^1(I, E)$. The cleanest way to see this is through a linear operator $L$ whose kernel is the homogeneous solution space $\mathcal{S}_0$, and whose preimage of $b$ is the full solution set $\mathcal{S}$.

Proposition — The solution operator

The map $$ L \colon \mathcal{C}^1(I, E) \to \mathcal{C}^0(I, E), \quad x \mapsto x' - a \cdot x $$ is well-defined and linear. The equation $(E)$ reads $L(x) = b$, and the homogeneous equation $(E_0)$ reads $L(x) = 0$.

Proof

For $x \in \mathcal{C}^1(I, E)$, the derivative $x'$ is continuous, $a$ is continuous, and the bilinear evaluation $(u, e) \mapsto u \cdot e$ is continuous on a finite-dimensional space, so $a \cdot x$ is continuous and $L(x) = x' - a \cdot x \in \mathcal{C}^0(I, E)$: $L$ is well-defined. Linearity follows from the linearity of the derivative and of $x \mapsto a \cdot x$: for $x_1, x_2 \in \mathcal{C}^1(I, E)$ and $\lambda, \mu \in \mathbb{K}$, $$ \begin{aligned} L(\lambda x_1 + \mu x_2) &= (\lambda x_1 + \mu x_2)' - a \cdot (\lambda x_1 + \mu x_2) && \\ &= \lambda x_1' + \mu x_2' - \lambda(a \cdot x_1) - \mu(a \cdot x_2) && \text{(linearity of $'$ and of $a \cdot$)} \\ &= \lambda L(x_1) + \mu L(x_2). && \end{aligned} $$

Theorem — Structure of the solution set

The set $\mathcal{S}_0 = \ker L$ of solutions of $(E_0)$ is a linear subspace of $\mathcal{C}^1(I, E)$. The linear Cauchy theorem of \S 1.2 guarantees that $(E)$ has at least one solution $\varphi_p$, and the set $\mathcal{S}$ of solutions of $(E)$ is then the affine subspace $$ \mathcal{S} = \varphi_p + \mathcal{S}_0. $$

Proof

$\mathcal{S}_0$ is the kernel of the linear map $L$, hence a linear subspace of $\mathcal{C}^1(I, E)$. The existence of a particular solution $\varphi_p$ is the linear Cauchy theorem applied to any initial condition (e.g. $(t_0, 0)$ for any fixed $t_0 \in I$). For the affine description: a function $\varphi$ solves $(E)$ iff $L(\varphi) = b = L(\varphi_p)$, iff $L(\varphi - \varphi_p) = 0$, iff $\varphi - \varphi_p \in \mathcal{S}_0$, iff $\varphi \in \varphi_p + \mathcal{S}_0$.

Proposition — Superposition principle

Let $b_1, b_2 \in \mathcal{C}(I, E)$ and $\lambda_1, \lambda_2 \in \mathbb{K}$. If $\varphi_i$ is a solution of $x' = a(t) \cdot x + b_i(t)$ for $i = 1, 2$, then $\lambda_1 \varphi_1 + \lambda_2 \varphi_2$ is a solution of $x' = a(t) \cdot x + (\lambda_1 b_1 + \lambda_2 b_2)(t)$.

Proof

Linearity of $L$: $L(\lambda_1 \varphi_1 + \lambda_2 \varphi_2) = \lambda_1 L(\varphi_1) + \lambda_2 L(\varphi_2) = \lambda_1 b_1 + \lambda_2 b_2$.

Theorem — Dimension of the solution space

Fix $t_0 \in I$. The evaluation map at $t_0$, $$ \delta_{t_0} \colon \mathcal{S}_0 \to E, \quad \varphi \mapsto \varphi(t_0), $$ is a linear isomorphism. In particular, $$ \dim \mathcal{S}_0 = \dim E. $$

Proof

$\delta_{t_0}$ is linear. For bijectivity, apply the linear Cauchy theorem of \S 1.2 with $b = 0$ (the homogeneous case is the special case of the inhomogeneous statement): for every $x_0 \in E$, there is one and only one $\varphi \in \mathcal{S}_0$ with $\varphi(t_0) = x_0$ --- existence gives surjectivity, uniqueness gives injectivity. Hence $\delta_{t_0}$ is an isomorphism of $\mathbb{K}$-vector spaces, and $\dim \mathcal{S}_0 = \dim E$.

Definition — Fundamental system of solutions

A fundamental system of solutions of $(E_0)$ is a basis $(\varphi_1, \ldots, \varphi_n)$ of the solution space $\mathcal{S}_0$, where $n = \dim E$.

Proposition — Characterisation of a fundamental system

Let $(\varphi_1, \ldots, \varphi_n)$ be a family of elements of $\mathcal{S}_0$ with $n = \dim E$. The following are equivalent:

[(i)] $(\varphi_1, \ldots, \varphi_n)$ is a fundamental system;
[(ii)] there exists $t_0 \in I$ such that $\bigl(\varphi_1(t_0), \ldots, \varphi_n(t_0)\bigr)$ is a basis of $E$;
[(iii)] for every $t \in I$, $\bigl(\varphi_1(t), \ldots, \varphi_n(t)\bigr)$ is a basis of $E$.

Proof

For any $t \in I$, the isomorphism $\delta_t \colon \mathcal{S}_0 \to E$ sends a basis to a basis (and conversely). So $(\varphi_1, \ldots, \varphi_n)$ is a basis of $\mathcal{S}_0$ iff $(\delta_t(\varphi_1), \ldots, \delta_t(\varphi_n)) = (\varphi_1(t), \ldots, \varphi_n(t))$ is a basis of $E$. This gives (i) $\iff$ (ii) (for any one $t$) and (i) $\iff$ (iii) (for every $t$), hence the three-way equivalence.

Example — A homogeneous planar system$\virgule$ revisited

For the planar system $x' = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \cdot x$ on $E = \mathbb{R}^2$, the dimension theorem gives $\dim \mathcal{S}_0 = 2$. The two solutions $\varphi_1 \colon t \mapsto (\cos t, -\sin t)$ and $\varphi_2 \colon t \mapsto (\sin t, \cos t)$ take at $t = 0$ the values $\varphi_1(0) = (1, 0)$ and $\varphi_2(0) = (0, 1)$ --- the canonical basis of $\mathbb{R}^2$. By the characterisation above, $(\varphi_1, \varphi_2)$ is a fundamental system, and the general solution of the system is the linear combination $t \mapsto \lambda_1 \varphi_1(t) + \lambda_2 \varphi_2(t)$ with $\lambda_1, \lambda_2 \in \mathbb{R}$.

The geometric picture of the linear Cauchy theorem is striking: in the strip $I \times E$, the integral curves of $(E)$ form a foliation --- through each point passes exactly one curve, and two distinct curves never meet.

Through the point $(t_0, x_0)$ passes exactly one integral curve; two distinct curves never cross.

Method — Show that a family is a fundamental system

To prove that $(\varphi_1, \ldots, \varphi_n)$ with $n = \dim E$ is a fundamental system of $(E_0)$:

exhibit the $n$ functions and check that each one solves $(E_0)$;
pick any one convenient $t_0 \in I$ and verify that $\bigl(\varphi_1(t_0), \ldots, \varphi_n(t_0)\bigr)$ is a basis of $E$ (e.g. by computing a determinant in a basis of $E$).

Since $\dim \mathcal{S}_0 = n$, an alternative is to exhibit $n$ solutions and verify that they form a free family of cardinal $n$ in $\mathcal{S}_0$: a free family of the right cardinal in an $n$-dimensional space is a basis.

Skills to practice

Using the structure of the solution set

I.4 Matrix systems and the order-two scalar equation as a planar system

Two reformulations of $(E)$ are useful and concrete. The first: a basis of $E$ turns the abstract equation into a matrix system $X' = A(t)X + B(t)$. The second: a scalar equation of order two becomes a planar first-order system, through its companion matrix. The second case is the bridge to \S 2.

Definition — Linear differential system

Let $n \ge 1$, $A \in \mathcal{C}(I, \mathcal{M}_n(\mathbb{K}))$ and $B \in \mathcal{C}(I, \mathbb{K}^n)$. The linear differential system associated with $A$ and $B$ is $$ (\Sigma)\colon \quad X' = A(t)\,X + B(t), $$ where the unknown is a $\mathcal{C}^1$ column-valued map $X \colon I \to \mathbb{K}^n$.

Example — Reading a system off its equations

On $I = \mathbb{R}$ and $n = 2$, the system $$ \begin{cases} x_1' = 2\,x_1 + t\,x_2 + 1, \\ x_2' = x_1 + \cos t \end{cases} $$ fits the Definition with $$ A(t) = \begin{pmatrix} 2 & t \\ 1 & 0 \end{pmatrix}, \qquad B(t) = \begin{pmatrix} 1 \\ \cos t \end{pmatrix}, $$ both continuous on $\mathbb{R}$. The unknown column $X = (x_1, x_2)$ is to be found in $\mathcal{C}^1(\mathbb{R}, \mathbb{R}^2)$. The matrix $A$ is non-constant (entry $a_{12}(t) = t$ depends on $t$), so this is a genuine variable-coefficient system; the linear Cauchy theorem of \S 1.2, transported through the Proposition below, will produce one and only one solution per initial condition $X(t_0) = X_0 \in \mathbb{R}^2$.

Proposition — Equation $\Leftrightarrow$ system

Fix a basis $\mathcal{B}$ of $E$. A map $\varphi \colon I \to E$ solves $(E)\colon x' = a(t) \cdot x + b(t)$ if and only if its coordinate column $X \colon I \to \mathbb{K}^n$ in $\mathcal{B}$ solves $$ X' = A(t)\,X + B(t), $$ where $A(t)$ is the matrix of $a(t)$ in $\mathcal{B}$ and $B(t)$ the coordinate column of $b(t)$ in $\mathcal{B}$.

Proof

Reading $(E)$ coordinatewise in $\mathcal{B}$: for $\varphi(t) = \sum_j X_j(t)\,e_j$, the derivative is $\varphi'(t) = \sum_j X_j'(t)\,e_j$. The right-hand side $a(t) \cdot \varphi(t) + b(t)$ has coordinate column $A(t)\,X(t) + B(t)$. The equation $\varphi'(t) = a(t) \cdot \varphi(t) + b(t)$ holds for every $t$ iff the coordinate columns agree at every $t$, iff $X'(t) = A(t)\,X(t) + B(t)$.

Proposition — Order-two scalar equation $\Leftrightarrow$ planar first-order system

Let $a_0, a_1, b \in \mathcal{C}(I, \mathbb{K})$. The scalar order-two equation $$ x'' + a_1(t)\,x' + a_0(t)\,x = b(t) $$ and the planar first-order system $$ X' = A(t)\,X + B(t) \quad \text{with} \quad A(t) = \begin{pmatrix} 0 & 1 \\ -a_0(t) & -a_1(t) \end{pmatrix}, \quad B(t) = \begin{pmatrix} 0 \\ b(t) \end{pmatrix} $$ are equivalent in the following sense:

[(i)] if $\varphi$ is a $\mathcal{C}^2$ solution of the scalar equation, then $\Phi = (\varphi, \varphi')$ is a $\mathcal{C}^1$ solution of the planar system;
[(ii)] conversely, every $\mathcal{C}^1$ solution $X = (u, v)$ of the planar system automatically satisfies $v = u'$ (read from the first row), hence $u$ is $\mathcal{C}^2$ and solves the scalar equation.

The correspondence $\varphi \leftrightarrow (\varphi, \varphi')$ is a bijection between the two solution sets.

Proof

(i). Let $\varphi$ be a $\mathcal{C}^2$ solution of the scalar equation. Set $\Phi = (\varphi, \varphi')$. Then $\Phi$ is $\mathcal{C}^1$ (its two components are $\mathcal{C}^2$ and $\mathcal{C}^1$ respectively, both at least $\mathcal{C}^1$). The first component of $\Phi'$ is $\varphi'$, which equals the first component of $A(t)\,\Phi(t) + B(t)$, namely $0 \cdot \varphi + 1 \cdot \varphi' + 0 = \varphi'$. The second component of $\Phi'$ is $\varphi''$, which by the scalar equation equals $-a_0(t)\,\varphi - a_1(t)\,\varphi' + b(t)$ --- exactly the second component of $A(t)\,\Phi(t) + B(t)$.
(ii). Conversely, let $X = (u, v)$ be a $\mathcal{C}^1$ solution of the planar system. The first row of $X' = A(t)\,X + B(t)$ reads $u' = v$. Since $X$ is $\mathcal{C}^1$, its component $v$ is $\mathcal{C}^1$; and $u' = v$, so $u'$ is $\mathcal{C}^1$, hence $u$ is $\mathcal{C}^2$. The second row reads $v' = -a_0(t)\,u - a_1(t)\,v + b(t)$; substituting $v = u'$ gives $u'' = -a_0(t)\,u - a_1(t)\,u' + b(t)$, i.e. $u'' + a_1(t)\,u' + a_0(t)\,u = b(t)$ --- the scalar equation.

Example — The harmonic oscillator as a planar system

The scalar equation $x'' + \omega^2 x = 0$ (with $\omega \in \mathbb{R}^*$ constant) is the order-two case of the Proposition, with $a_1 = 0$, $a_0 = \omega^2$, $b = 0$. Its companion system reads $$ X' = \begin{pmatrix} 0 & 1 \\ -\omega^2 & 0 \end{pmatrix} \cdot X. $$ The structure theorems of \S 1.3 applied to this planar system give $\dim \mathcal{S}_0 = 2$, with the constant-coefficient resolution (MPSI) furnishing the basis $(\cos(\omega t), \sin(\omega t))$ of scalar solutions, equivalently $((\cos(\omega t), -\omega \sin(\omega t)), (\sin(\omega t), \omega \cos(\omega t)))$ in the planar form.

Method — Reduce a scalar order-two equation to a planar system

Given a scalar order-two equation $x'' + a_1(t)\,x' + a_0(t)\,x = b(t)$, set $X = (x, x')$. The Proposition above turns the order-two problem into the planar first-order system $$ X' = \begin{pmatrix} 0 & 1 \\ -a_0(t) & -a_1(t) \end{pmatrix}\,X + \begin{pmatrix} 0 \\ b(t) \end{pmatrix}, $$ to which the general theory of \S 1.2--1.3 applies with $\dim E = 2$. The regularity-promotion step (ii) of the Proposition is the bridge that turns a $\mathcal{C}^1$ planar solution into a $\mathcal{C}^2$ scalar one.

Skills to practice

Reducing an order-two equation to a planar system

II Scalar linear equations of order two

II.1 The order-two equation and its solution space

The case the program singles out. The coefficients are now allowed to vary with $t$ --- a strict generalisation of the constant-coefficient case treated in MPSI. The solution space of the homogeneous equation is a plane in $\mathcal{C}^2(I, \mathbb{K})$, and the full solution set of the inhomogeneous equation is a plane translated by one particular solution.

Definition — Scalar linear equation of order two

Let $a_0, a_1, b \in \mathcal{C}(I, \mathbb{K})$. The normalised scalar linear equation of order two is $$ (E)\colon \quad x'' + a_1(t)\,x' + a_0(t)\,x = b(t). $$ A solution of $(E)$ on $I$ is a map $x \colon I \to \mathbb{K}$ of class $\mathcal{C}^2$ satisfying $(E)$. The homogeneous companion is $$ (E_0)\colon \quad x'' + a_1(t)\,x' + a_0(t)\,x = 0. $$

Theorem — Cauchy theorem and structure$\virgule$ order two

For every $(t_0, x_0, x_1) \in I \times \mathbb{K} \times \mathbb{K}$, the Cauchy problem $$ \begin{cases} x'' + a_1(t)\,x' + a_0(t)\,x = b(t), \\ x(t_0) = x_0, \quad x'(t_0) = x_1 \end{cases} $$ admits one and only one solution, defined on all of $I$. The solution space $\mathcal{S}_0$ of $(E_0)$ has dimension $2$ over $\mathbb{K}$ (a $\mathbb{K}$-plane), and the solution set of $(E)$ is the affine $\mathbb{K}$-plane $\mathcal{S} = \varphi_p + \mathcal{S}_0$ for any particular solution $\varphi_p$.

Proof

The Proposition of \S 1.4 turns the order-two Cauchy problem into the planar Cauchy problem $$ \begin{cases} X' = A(t)\,X + B(t), \\ X(t_0) = (x_0, x_1) \end{cases} $$ with the companion matrix. By the linear Cauchy theorem of \S 1.2 (applied to the planar system on $E = \mathbb{K}^2$), this planar problem has a unique $\mathcal{C}^1$ solution $X = (u, v)$ on $I$. By the regularity-promotion step (ii) of the \S 1.4 Proposition, $u$ is automatically $\mathcal{C}^2$ and solves the scalar equation with $u(t_0) = x_0$, $u'(t_0) = v(t_0) = x_1$ --- existence and uniqueness for the scalar problem follow. For the structure, the bijection $\varphi \leftrightarrow (\varphi, \varphi')$ identifies the scalar $\mathcal{S}_0$ with the planar one, and the planar one has dimension $\dim \mathbb{K}^2 = 2$ by the \S 1.3 dimension theorem. The affine description $\mathcal{S} = \varphi_p + \mathcal{S}_0$ is the \S 1.3 structure theorem.

Proposition — Scalar order-two evaluation isomorphism

Fix $t_0 \in I$. The evaluation map $$ \delta_{t_0} \colon \mathcal{S}_0 \to \mathbb{K}^2, \quad \varphi \mapsto \bigl(\varphi(t_0) \,;\, \varphi'(t_0)\bigr) $$ is a linear isomorphism of $\mathbb{K}$-vector spaces. In particular $\dim \mathcal{S}_0 = 2$.

Proof

Linearity of $\delta_{t_0}$ is direct. Bijectivity is the Cauchy theorem just established: for every pair $(x_0, x_1) \in \mathbb{K}^2$, there is one and only one $\varphi \in \mathcal{S}_0$ with $\varphi(t_0) = x_0$, $\varphi'(t_0) = x_1$.

Example — The constant-coefficient case recovered

For $a_1 = 0$ and $a_0 = 1$ constant, the homogeneous equation is $x'' + x = 0$. The MPSI resolution gives $\mathcal{S}_0 = \mathrm{Vect}(\cos, \sin)$. Indeed both $\cos$ and $\sin$ are $\mathcal{C}^2$ and satisfy the equation; the pair $(\cos, \sin)$ takes at $t = 0$ the values $(\cos 0, \cos'(0)) = (1, 0)$ and $(\sin 0, \sin'(0)) = (0, 1)$ --- the canonical basis of $\mathbb{K}^2$. By the characterisation of \S 1.3 (transported through the order-two evaluation isomorphism), $(\cos, \sin)$ is a fundamental system of $(E_0)$.

Proposition — Isolated zeros

Let $\varphi$ be a non-zero solution of $(E_0)$ on $I$. The zeros of $\varphi$ are isolated: for every $t_0 \in I$ with $\varphi(t_0) = 0$, there is a neighbourhood of $t_0$ in $I$ (two-sided if $t_0$ is interior to $I$, one-sided if $t_0$ is an endpoint) on which $\varphi$ vanishes only at $t_0$.

Proof

At a zero $t_0$ of $\varphi$, $\varphi'(t_0) \ne 0$ --- otherwise the Cauchy data $(\varphi(t_0), \varphi'(t_0)) = (0, 0)$ would (by uniqueness) force $\varphi \equiv 0$ on $I$, contradicting the hypothesis. The Taylor-Young expansion of $\varphi$ at $t_0$ as a $\mathbb{K}$-valued function of the real variable $t$ reads $\varphi(t) = \varphi'(t_0)\,(t - t_0) + (t - t_0)\,\varepsilon(t)$ with $\varepsilon(t) \to 0$ as $t \to t_0$. For $|t - t_0|$ small enough, $|\varepsilon(t)| < |\varphi'(t_0)|/2$, hence $$ \left| \frac{\varphi(t)}{t - t_0} \right| \ge |\varphi'(t_0)| - |\varepsilon(t)| > \frac{|\varphi'(t_0)|}{2} > 0, $$ so $\varphi(t) \ne 0$ on a punctured neighbourhood of $t_0$ in $I$. The argument is valid in $\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$.

Method — Frame an order-two problem

Faced with a scalar order-two equation $x'' + a_1(t)\,x' + a_0(t)\,x = b(t)$:

identify $a_0, a_1, b$ and the working interval $I$ on which they are continuous;
by the theorem above, the homogeneous solution set $\mathcal{S}_0$ is a $\mathbb{K}$-plane, and the full solution set $\mathcal{S}$ is a plane translated by one particular solution $\varphi_p$;
the search for a fundamental system of $(E_0)$ and for one particular solution of $(E)$ is then the entire task --- this is what \S 2.2--2.4 makes operational.

Skills to practice

Framing the solution space of an order-two equation

II.2 The wronskian

A two-by-two determinant that detects whether two solutions of $(E_0)$ form a basis of the solution plane. Its key property: as a function of $t$, the wronskian solves a scalar order-one equation, so it is either identically zero or never zero on $I$.

Definition — Wronskian

Let $\varphi_1, \varphi_2 \colon I \to \mathbb{K}$ be derivable. The wronskian of the pair $(\varphi_1, \varphi_2)$ is the function $$ W_{\varphi_1, \varphi_2} \colon I \to \mathbb{K}, \quad t \mapsto \det \begin{pmatrix} \varphi_1(t) & \varphi_2(t) \\ \varphi_1'(t) & \varphi_2'(t) \end{pmatrix} = \varphi_1(t)\,\varphi_2'(t) - \varphi_1'(t)\,\varphi_2(t). $$

Proposition — Differential equation of the wronskian

Let $(\varphi_1, \varphi_2) \in \mathcal{S}_0^2$ be a pair of solutions of $(E_0)$. The wronskian $W = W_{\varphi_1, \varphi_2}$ satisfies $$ W'(t) = -a_1(t)\,W(t) \quad \text{on } I. $$ Hence $W = \lambda \exp(-A_1)$ for some constant $\lambda \in \mathbb{K}$, where $A_1$ is a primitive of $a_1$. In particular $W$ is either identically zero or never zero on $I$.

Proof

Differentiate $W = \varphi_1 \varphi_2' - \varphi_1' \varphi_2$. Both $\varphi_i$ are $\mathcal{C}^2$, so $W$ is derivable: $$ \begin{aligned} W'(t) &= \varphi_1'\varphi_2' + \varphi_1\varphi_2'' - \varphi_1''\varphi_2 - \varphi_1'\varphi_2' && \\ &= \varphi_1\varphi_2'' - \varphi_1''\varphi_2. && \end{aligned} $$ Substitute $\varphi_i'' = -a_1\,\varphi_i' - a_0\,\varphi_i$ (since $\varphi_i \in \mathcal{S}_0$): $$ \begin{aligned} W'(t) &= \varphi_1\bigl(-a_1\varphi_2' - a_0\varphi_2\bigr) - \bigl(-a_1\varphi_1' - a_0\varphi_1\bigr)\varphi_2 && \\ &= -a_1\bigl(\varphi_1\varphi_2' - \varphi_1'\varphi_2\bigr) - a_0\bigl(\varphi_1\varphi_2 - \varphi_1\varphi_2\bigr) && \\ &= -a_1\,W. && \end{aligned} $$ The equation $W' = -a_1 W$ is a scalar order-one linear homogeneous equation; by the MPSI explicit formula, its general solution is $W(t) = \lambda \exp\!\bigl(-A_1(t)\bigr)$ with $\lambda \in \mathbb{K}$ and $A_1$ a primitive of $a_1$. Since $\exp \ne 0$, $W$ is either identically zero ($\lambda = 0$) or never zero ($\lambda \ne 0$).

Proposition — The wronskian detects a fundamental system

Let $(\varphi_1, \varphi_2) \in \mathcal{S}_0^2$. The following are equivalent:

[(i)] $(\varphi_1, \varphi_2)$ is a fundamental system of $(E_0)$;
[(ii)] there exists $t_0 \in I$ with $W(t_0) \ne 0$;
[(iii)] for every $t \in I$, $W(t) \ne 0$.

Proof

(ii) $\iff$ (iii). This is the dichotomy of the previous Proposition: the wronskian $W$ is either identically zero or never zero on $I$, so a non-vanishing at one point $t_0$ propagates to every point, and conversely.
(i) $\iff$ (ii). Fix $t_0 \in I$. By the order-two evaluation isomorphism $\delta_{t_0} \colon \mathcal{S}_0 \to \mathbb{K}^2$, $\varphi \mapsto (\varphi(t_0), \varphi'(t_0))$, the pair $(\varphi_1, \varphi_2)$ is a basis of $\mathcal{S}_0$ iff $(\delta_{t_0}(\varphi_1), \delta_{t_0}(\varphi_2)) = ((\varphi_1(t_0), \varphi_1'(t_0)), (\varphi_2(t_0), \varphi_2'(t_0)))$ is a basis of $\mathbb{K}^2$, iff the determinant $$ \det \begin{pmatrix} \varphi_1(t_0) & \varphi_2(t_0) \\ \varphi_1'(t_0) & \varphi_2'(t_0) \end{pmatrix} = W(t_0) $$ is non-zero.

Example — The wronskian of $\cos$ and $\sin$

For the equation $x'' + x = 0$, the pair $(\cos, \sin)$ has wronskian $$ W_{\cos, \sin}(t) = \cos t \cdot \cos t - (-\sin t) \cdot \sin t = \cos^2 t + \sin^2 t = 1. $$ This is non-zero (constant equal to $1$), so $(\cos, \sin)$ is a fundamental system. The constancy of $W$ is consistent with the Proposition: here $a_1 = 0$, so $W' = -a_1 W = 0$.

Example — When $a_1 \equal 0$$\virgule$ the wronskian is constant

For an equation of the form $x'' + a_0(t)\,x = 0$ (no first-derivative term, $a_1 \equiv 0$): for any pair $(\varphi_1, \varphi_2) \in \mathcal{S}_0^2$, $W' = -a_1 W = 0$, so $W$ is constant on $I$. To check whether $(\varphi_1, \varphi_2)$ is a fundamental system, evaluate $W$ at one convenient $t_0$ --- the value at one point is the value at every point.

Method — Test a fundamental system by the wronskian

Given two known solutions $\varphi_1, \varphi_2 \in \mathcal{S}_0$ of $(E_0)$, compute the wronskian $$ W(t) = \varphi_1(t)\,\varphi_2'(t) - \varphi_1'(t)\,\varphi_2(t) $$ at one convenient point $t_0 \in I$. If $W(t_0) \ne 0$: $(\varphi_1, \varphi_2)$ is a fundamental system of $(E_0)$, and the homogeneous general solution is $\lambda_1 \varphi_1 + \lambda_2 \varphi_2$ with $\lambda_1, \lambda_2 \in \mathbb{K}$. If $W(t_0) = 0$: $W$ is identically zero on $I$, and $(\varphi_1, \varphi_2)$ is linearly dependent.

Skills to practice

Computing a wronskian

II.3 The method of variation of constants

A fundamental system of the homogeneous equation $(E_0)$ being known, the method of variation of constants produces a particular solution of the complete equation $(E)$. The trick of the trade: replace the constants $\lambda_1, \lambda_2$ of the homogeneous combination by functions of $t$, and impose a convenient relation between them.

Theorem — Variation of constants$\virgule$ order two

Let $(\varphi_1, \varphi_2)$ be a fundamental system of $(E_0)$ with wronskian $W$ (non-zero on $I$). The Cramer system $$ \begin{cases} \lambda_1'\,\varphi_1 + \lambda_2'\,\varphi_2 = 0, \\ \lambda_1'\,\varphi_1' + \lambda_2'\,\varphi_2' = b \end{cases} $$ has determinant $W \ne 0$, and its unique solution is $$ \lambda_1' = -\frac{b\,\varphi_2}{W}, \quad \lambda_2' = \frac{b\,\varphi_1}{W}. $$ Both right-hand sides are continuous on $I$ (quotients of continuous functions with non-vanishing denominator), so any pair $(\lambda_1, \lambda_2)$ of primitives is of class $\mathcal{C}^1$ on $I$; the map $$ \varphi = \lambda_1\,\varphi_1 + \lambda_2\,\varphi_2 $$ is then $\mathcal{C}^2$ and is a particular solution of $(E)$. Two choices of primitives differ by an element of $\mathcal{S}_0$, so they yield the same affine solution set $\mathcal{S} = \varphi + \mathcal{S}_0$.

Proof

Set $\varphi = \lambda_1\varphi_1 + \lambda_2\varphi_2$ with $\lambda_1, \lambda_2$ derivable. Differentiate: $$ \varphi' = \lambda_1'\varphi_1 + \lambda_2'\varphi_2 + \lambda_1\varphi_1' + \lambda_2\varphi_2'. $$ Impose the first Cramer relation $\lambda_1'\varphi_1 + \lambda_2'\varphi_2 = 0$. Then $\varphi' = \lambda_1\varphi_1' + \lambda_2\varphi_2'$. Differentiate again: $$ \varphi'' = \lambda_1'\varphi_1' + \lambda_2'\varphi_2' + \lambda_1\varphi_1'' + \lambda_2\varphi_2''. $$ Substitute into $(E)$: $$ \begin{aligned} \varphi'' + a_1\varphi' + a_0\varphi &= \lambda_1'\varphi_1' + \lambda_2'\varphi_2' + \lambda_1\bigl(\varphi_1'' + a_1\varphi_1' + a_0\varphi_1\bigr) + \lambda_2\bigl(\varphi_2'' + a_1\varphi_2' + a_0\varphi_2\bigr) && \\ &= \lambda_1'\varphi_1' + \lambda_2'\varphi_2', && \text{(both $\varphi_i \in \mathcal{S}_0$)} \end{aligned} $$ which equals $b$ iff the second Cramer relation $\lambda_1'\varphi_1' + \lambda_2'\varphi_2' = b$ holds. The Cramer system has determinant $W \ne 0$ and solution $\lambda_1' = -b\varphi_2/W$, $\lambda_2' = b\varphi_1/W$. The continuity argument for the regularity of $\varphi$ is as stated.

Example — The tangent right-hand side $x'' + x \equal \tan t$

Take $a_1 = 0$, $a_0 = 1$, $b(t) = \tan t$ on $I = \,]{-}\pi/2 \,;\, \pi/2[$. The homogeneous equation $x'' + x = 0$ admits $(\cos, \sin)$ as a fundamental system (Example above), with wronskian $W = 1$. The Cramer formulas give $$ \lambda_1'(t) = -\frac{\tan t \cdot \sin t}{1} = -\frac{\sin^2 t}{\cos t}, \qquad \lambda_2'(t) = \frac{\tan t \cdot \cos t}{1} = \sin t. $$ A primitive of $\lambda_2'$ is $\lambda_2(t) = -\cos t$. For $\lambda_1'$, compute $\sin^2 t / \cos t = (1 - \cos^2 t)/\cos t = 1/\cos t - \cos t$. A primitive of $1/\cos t$ on $\,]{-}\pi/2 \,;\, \pi/2[$ is $t \mapsto \ln\!\bigl((1+\sin t)/\cos t\bigr)$ (the argument is positive there; classical, recalled from MPSI), so a primitive of $\sin^2 t / \cos t$ is $\ln\!\bigl((1+\sin t)/\cos t\bigr) - \sin t$, and consequently a primitive of $\lambda_1'(t) = -\sin^2 t/\cos t$ is $\lambda_1(t) = -\ln\!\bigl((1+\sin t)/\cos t\bigr) + \sin t$. A particular solution is therefore $$ \varphi(t) = \lambda_1(t)\,\cos t + \lambda_2(t)\,\sin t = -\cos t \cdot \ln\!\left(\frac{1 + \sin t}{\cos t}\right) $$ on $\,]{-}\pi/2 \,;\, \pi/2[$ (the $\sin t \cos t$ terms from $\lambda_1$ and $\lambda_2$ cancel).

Example — A Cauchy-Euler equation with right-hand side

Take $t^2 x'' - 2 t x' + 2 x = t^3$ on $I = \,]0 \,;\, +\infty[$. Divide by $t^2$ (legitimate since $t > 0$) to normalise: $x'' - (2/t)\,x' + (2/t^2)\,x = t$. So $a_1 = -2/t$, $a_0 = 2/t^2$, $b = t$. The homogeneous equation $t^2 x'' - 2 t x' + 2 x = 0$ admits power solutions $x = t^a$ with $a(a-1) - 2a + 2 = 0$, i.e. $a^2 - 3a + 2 = 0$, $a \in \{1, 2\}$. So $\varphi_1(t) = t$ and $\varphi_2(t) = t^2$ are solutions; the wronskian is $$ W = \varphi_1 \varphi_2' - \varphi_1' \varphi_2 = t \cdot 2t - 1 \cdot t^2 = t^2 \ne 0 \quad \text{on } I, $$ so $(\varphi_1, \varphi_2)$ is a fundamental system. With $b(t) = t$, the Cramer formulas give $$ \lambda_1'(t) = -\frac{t \cdot t^2}{t^2} = -t, \qquad \lambda_2'(t) = \frac{t \cdot t}{t^2} = 1. $$ Primitives: $\lambda_1(t) = -t^2/2$, $\lambda_2(t) = t$. A particular solution is therefore $$ \varphi_p(t) = \lambda_1\,\varphi_1 + \lambda_2\,\varphi_2 = -\frac{t^2}{2} \cdot t + t \cdot t^2 = \frac{t^3}{2}, $$ and the general solution of the complete equation is $t \mapsto t^3/2 + \alpha t + \beta t^2$ with $\alpha, \beta \in \mathbb{K}$.

Method — Apply variation of constants

Given $(E)\colon x'' + a_1(t)\,x' + a_0(t)\,x = b(t)$ on $I$:

[(1)] find a fundamental system $(\varphi_1, \varphi_2)$ of $(E_0)$ (by inspection, by the constant-coefficient resolution of MPSI, by one of the techniques of \S 2.4, \ldots);
[(2)] compute the wronskian $W$ at one point to confirm it is non-zero --- the dichotomy then ensures $W \ne 0$ on all of $I$;
[(3)] solve the Cramer system: $\lambda_1' = -b\,\varphi_2/W$, $\lambda_2' = b\,\varphi_1/W$;
[(4)] primitivate $\lambda_1'$ and $\lambda_2'$ to obtain $\lambda_1$ and $\lambda_2$ (any choice of primitives works);
[(5)] the particular solution is $\varphi_p = \lambda_1 \varphi_1 + \lambda_2 \varphi_2$, and the general solution of $(E)$ is $\mathcal{S} = \varphi_p + \mathcal{S}_0$.

Skills to practice

Applying variation of constants

II.4 Solving techniques

When no fundamental system of $(E_0)$ is handed to you, three worked-example techniques for the order-two equation --- applied pragmatically on a case-by-case basis, not stated as general theorems: a polynomial search (when the coefficients are polynomial), reduction of order from one known non-vanishing solution, and a change of variable that brings the equation to constant coefficients.

Method — Look for polynomial solutions --- a worked-example heuristic

On a normalised order-two equation with polynomial coefficients, inject an ansatz $P(t) = \sum_{k=0}^{d} a_k\,t^k$ of provisional degree $d$, with $a_d \ne 0$. Expand and identify coefficients of $t^k$: a finite linear system on $(a_k)$ results. Two outcomes:

the system fixes $d$ via a leading-term degree argument, and the remaining coefficients are determined by a finite recurrence --- the polynomial solutions are then explicitly known;
or the leading-term comparison is incompatible with $a_d \ne 0$, and no polynomial solution of that degree exists --- the analysis can then be re-run at any other candidate degree.

The technique is a heuristic applied case by case, not a blanket theorem.

Example — A polynomial solution of $x'' - 2 t x' + 2 x \equal 0$

Take the normalised equation $x'' - 2 t x' + 2 x = 0$ on $\mathbb{R}$. Inject a polynomial $P(t) = \sum_{k=0}^{d} a_k t^k$ of degree $d \ge 0$ with $a_d \ne 0$. The highest-degree contribution of $P'' - 2 t P' + 2 P$ is the degree-$d$ term: $P''$ contributes degree $d - 2$ (negligible), $-2 t P'$ contributes $-2 d\,a_d\,t^d$, and $2 P$ contributes $2\,a_d\,t^d$. The leading term is therefore $2(1 - d)\,a_d\,t^d$. For $P$ to solve the equation, this must vanish; with $a_d \ne 0$, the only option is $d = 1$.
So $P(t) = a_0 + a_1 t$ with $a_1 \ne 0$. Substituting: $$ P'' - 2 t P' + 2 P = 0 - 2 t\,a_1 + 2(a_0 + a_1 t) = 2\,a_0, $$ which vanishes iff $a_0 = 0$. The polynomial solutions are exactly the scalar multiples of $\varphi(t) = t$. (This is the $n = 1$ instance of Hermite's family $x'' - 2 t x' + 2 n x = 0$ for $n \in \mathbb{N}$, whose polynomial solutions are the Hermite polynomials of degree $n$ --- a classical pattern.)

Method — Reduce the order knowing one solution

The prépa-tradition substitution-and-cancel route. Given a known $\varphi_1 \in \mathcal{S}_0$ non-vanishing on a subinterval $J \subset I$, seek a second solution of the form $\varphi = \varphi_1 \cdot z$ with $z \colon J \to \mathbb{K}$ to be found. Compute by Leibniz: $$ \varphi' = \varphi_1'\,z + \varphi_1\,z', \quad \varphi'' = \varphi_1''\,z + 2\,\varphi_1'\,z' + \varphi_1\,z''. $$ Substitute into $(E_0)$ and collect the coefficient of $z$: $\varphi_1'' + a_1\,\varphi_1' + a_0\,\varphi_1 = 0$ since $\varphi_1 \in \mathcal{S}_0$, so the $z$-term cancels. What remains is $$ \varphi_1\,z'' + (2\,\varphi_1' + a_1\,\varphi_1)\,z' = 0, $$ a first-order linear equation in the unknown $w = z'$. Solve it (by separation of variables or by an integrating factor), then primitivate $w$ to obtain $z$, and form $\varphi_2 = \varphi_1 \cdot z$. The classical pitfall is to stop at $z$ itself --- $z$ is not the new solution; the multiplication by $\varphi_1$ is required.

Example — Reduce the order on $x'' + x'/t - x/t^2 \equal 0$

On $I = \,]0 \,;\, +\infty[$, consider $x'' + x'/t - x/t^2 = 0$. The function $\varphi_1(t) = t$ is a solution (check: $0 + 1/t - 1/t = 0$). Set $\varphi = t\,z$. Then $\varphi' = z + t\,z'$, $\varphi'' = 2\,z' + t\,z''$. Substitute: $$ \begin{aligned} \varphi'' + \frac{\varphi'}{t} - \frac{\varphi}{t^2} &= 2\,z' + t\,z'' + \frac{z + t\,z'}{t} - \frac{t\,z}{t^2} && \\ &= t\,z'' + 2\,z' + \frac{z}{t} + z' - \frac{z}{t} && \\ &= t\,z'' + 3\,z'. && \end{aligned} $$ The equation reads $t\,z'' + 3\,z' = 0$, i.e. $w' + (3/t)\,w = 0$ for $w = z'$. The MPSI explicit formula gives $w(t) = C/t^3$ with $C \in \mathbb{K}$. Primitivating, $z(t) = -C/(2 t^2) + D$ with $D \in \mathbb{K}$. Picking $(C, D) = (-2, 0)$ yields $z(t) = 1/t^2$, hence $\varphi_2(t) = t \cdot 1/t^2 = 1/t$. This $\varphi_2$ is independent of $\varphi_1 = t$, so $(\varphi_1, \varphi_2) = (t, 1/t)$ is a fundamental system on $\,]0 \,;\, +\infty[$.

Method — Change of variable

Given a $\mathcal{C}^2$ map $\theta \colon J \to I$ between intervals that is bijective and has non-vanishing derivative $\theta'$ on $J$ (so $\theta^{-1}$ is automatically $\mathcal{C}^2$, making $\theta$ a $\mathcal{C}^2$-diffeomorphism), the substitution $u = \theta^{-1}(t)$ (equivalently $t = \theta(u)$) transports a solution $\varphi \colon I \to \mathbb{K}$ of $(E)$ to the function $\psi = \varphi \circ \theta \colon J \to \mathbb{K}$. The chain rule gives $$ \psi'(u) = \theta'(u)\,\varphi'(\theta(u)), \quad \psi''(u) = \theta''(u)\,\varphi'(\theta(u)) + \bigl(\theta'(u)\bigr)^2\,\varphi''(\theta(u)). $$ Solving for $\varphi', \varphi''$ in terms of $\psi', \psi''$ and substituting into $(E)$ yields an equivalent order-two equation in $\psi$ on $J$. When the new equation has constant coefficients, it is solved by MPSI's characteristic equation; the change is inverted at the end ($\varphi(t) = \psi(\theta^{-1}(t))$) to recover $\varphi$.

Example — $t^2 x'' + 3 t x' + 4 x \equal t \ln t$ by $t \equal e^u$

On $I = \,]0 \,;\, +\infty[$, dividing by $t^2$ (legitimate since $t > 0$) normalises the equation. Set $\theta(u) = e^u$, a $\mathcal{C}^\infty$-diffeomorphism $\mathbb{R} \to \,]0 \,;\, +\infty[$, and $\psi(u) = \varphi(e^u)$. The chain rule yields $$ \psi'(u) = e^u\,\varphi'(e^u) = t\,\varphi'(t), \quad \psi''(u) = e^{2 u}\,\varphi''(e^u) + e^u\,\varphi'(e^u) = t^2\,\varphi''(t) + t\,\varphi'(t), $$ hence $t^2\,\varphi''(t) = \psi''(u) - \psi'(u)$. The equation $t^2 \varphi'' + 3 t \varphi' + 4\varphi = t \ln t$ becomes $$ \bigl(\psi'' - \psi'\bigr) + 3\,\psi' + 4\,\psi = u\,e^u, \quad \text{i.e.} \quad \psi'' + 2\,\psi' + 4\,\psi = u\,e^u. $$ This is a constant-coefficient order-two equation in $u$, solved by MPSI's method. The characteristic equation $r^2 + 2 r + 4 = 0$ has complex conjugate roots $r = -1 \pm i \sqrt{3}$, so the homogeneous general solution is $\psi_h(u) = e^{-u}\bigl(\lambda \cos(\sqrt{3}\,u) + \mu \sin(\sqrt{3}\,u)\bigr)$ with $\lambda, \mu \in \mathbb{R}$. For a particular solution we try the ansatz $\psi_p(u) = (\alpha u + \beta)\,e^u$ (since $1$ is not a root of the characteristic equation); plugging in and identifying coefficients of $u e^u$ and $e^u$ gives $7\alpha = 1$ and $7\beta + 4\alpha = 0$, so $\alpha = 1/7$ and $\beta = -4/49$. Inverting via $u = \ln t$ recovers $$ \varphi(t) = \frac{1}{t}\bigl(\lambda \cos(\sqrt{3}\,\ln t) + \mu \sin(\sqrt{3}\,\ln t)\bigr) + \frac{t \ln t}{7} - \frac{4\,t}{49}, \qquad \lambda, \mu \in \mathbb{R}, $$ on $\,]0 \,;\, +\infty[$.

Scope note --- non-normalised equations

Non-normalised equations $\alpha(t)\,x'' + \beta(t)\,x' + \gamma(t)\,x = \delta(t)$ whose leading coefficient $\alpha$ vanishes at one or more points of $I$ lie outside this chapter's scope. The program (\S 10 e) frames the theory for normalised order-two equations on an interval $I$, and the gluing (raccord) of solution branches at a singular point is delicate and case-dependent --- neither a clean recipe nor a program objective. Recalled in passing from MPSI; not re-treated here.

Going further

This chapter built the general linear differential equation $x' = a(t) \cdot x + b(t)$ on a finite-dimensional space, its linear Cauchy theorem, the dim-$\dim E$ structure of its solution set, and the order-two specialisation with wronskian and variation of constants. The chapter Matrix exponential continues the story: when the coefficient $a$ is constant (in $\mathcal{L}(E)$, or equivalently $A$ constant in $\mathcal{M}_n(\mathbb{K})$), the homogeneous system $X' = A\,X$ is solved explicitly by the matrix exponential $X(t) = e^{(t - t_0)A}\,X_0$. The linear Cauchy theorem stated and admitted here is exactly what makes that resolution work; the structure and dimension theorems are exactly the framework in which $e^{tA}$ operates.

Skills to practice

Solving an order-two equation by a classical technique