Differential Equations Domain Guide

A linear differential operator behaves exactly like a matrix: the homogeneous solutions are its kernel, and solving a nonhomogeneous equation is solving $L y = g$ — one particular solution plus anything in the kernel. That is why an $n$-th order linear equation has exactly $n$ arbitrary constants, and why 'find the general solution' means 'find a basis for the kernel and one particular solution'.

Constant coefficients turn calculus into algebra. Substituting $y = e^{rx}$ converts the differential equation into a polynomial equation in $r$, so the whole solution structure — exponential growth, oscillation, resonance — is read from the roots. Systems are the same trick with $\vec{x} = e^{\lambda t}\vec{v}$, where the polynomial equation becomes the eigenvalue problem for $A$.

Many exam items never ask for a solution at all. They ask what happens as $t \to \infty$, or whether an equilibrium is stable. Those are sign questions: negative real parts pull solutions to zero, positive real parts push them away, and a scalar equilibrium is stable exactly when the slope of $f$ there is negative.

Classification comes first because it determines everything downstream. Write the order, then whether the equation is linear, then whether the coefficients are constant, then whether it is homogeneous. A first-order equation that is separable is solved by separating; one that is linear is solved by an integrating factor $\mu = e^{\int p}$ after normalising the leading coefficient to $1$; some equations are both, and either route works. Second order with constant coefficients goes to the characteristic polynomial. Anything else on this exam is usually recognisable as a Cauchy–Euler equation $x^2 y'' + bxy' + cy = 0$, which is solved by trying $y = x^r$.

Existence and uniqueness is examined mainly through its failure. For $y' = f(x,y)$ with $y(x_0) = y_0$, continuity of $f$ near the point gives a solution, and continuity of $\partial f / \partial y$ as well gives uniqueness. Remove the second hypothesis and uniqueness genuinely breaks: $y' = y^{2/3}$ with $y(0) = 0$ is solved both by $y \equiv 0$ and by $y = (x/3)^3$, because $\partial f/\partial y = \tfrac{2}{3}y^{-1/3}$ blows up at $y = 0$. Linear equations are better behaved — for $y' + p(x)y = q(x)$ with $p, q$ continuous on an interval, the solution through any point exists and is unique on that whole interval.

The structure theorem organises second-order work. For $ay'' + by' + cy = g(x)$, first solve $ay'' + by' + cy = 0$ via $ar^2 + br + c = 0$: distinct real roots $r_1 \ne r_2$ give $c_1 e^{r_1 x} + c_2 e^{r_2 x}$; a repeated root $r$ gives $(c_1 + c_2 x)e^{rx}$; complex roots $\alpha \pm \beta i$ give $e^{\alpha x}(c_1 \cos \beta x + c_2 \sin \beta x)$. Then add any particular solution of the full equation. Undetermined coefficients supplies that particular solution when $g$ is a polynomial, exponential, sine/cosine, or product of these — but if the trial form already solves the homogeneous equation, multiply it by $x$ (or $x^2$ for a repeated root). This resonance adjustment is one of the most reliably tested points in the domain.

Independence of solutions is checked with the Wronskian $W = y_1y_2' - y_1'y_2$. If $W \ne 0$ at even one point of the interval, the two solutions are independent and together span the solution space; Abel's identity shows that for $y'' + p(x)y' + q(x)y = 0$ the Wronskian is $W(x) = W(x_0)e^{-\int p}$, so it is either identically zero or never zero — which is why a single evaluation settles the question.

Systems are where linear algebra re-enters. For $\vec{x}' = A\vec{x}$, substituting $\vec{x} = e^{\lambda t}\vec{v}$ gives $(A - \lambda I)\vec{v} = \vec{0}$, so $\lambda$ must be an eigenvalue and $\vec{v}$ a corresponding eigenvector. With $n$ independent eigenvectors the general solution is $\sum c_i e^{\lambda_i t}\vec{v}_i$. Complex eigenvalues $\alpha \pm \beta i$ produce spirals: real solutions are $e^{\alpha t}$ times sines and cosines of $\beta t$. A repeated eigenvalue with too few eigenvectors produces a generalised eigenvector and a solution containing $t e^{\lambda t}$ — the exact analogue of the repeated-root factor of $x$ in the scalar case.

Qualitative behaviour is often the whole question. For a system, if every eigenvalue has negative real part all solutions decay to the origin (asymptotically stable); if any has positive real part the origin is unstable; one positive and one negative gives a saddle, and purely imaginary eigenvalues give closed orbits around a centre. For a scalar autonomous equation $y' = f(y)$, the equilibria are the roots of $f$, and an equilibrium $y^*$ is stable when $f'(y^*) < 0$ and unstable when $f'(y^*) > 0$ — the sign chart of $f$ is a complete phase line. The logistic equation $y' = ry(1 - y/K)$ with $r, K > 0$ illustrates both: $y = 0$ is unstable and $y = K$ is stable, so every solution starting positive tends to $K$ without any formula being needed.