differential equation first-order linear কী?

Example 5.3 (A homogeneous equation with variable coefficient)

Solve $y' + (\cos x)\,y = 0$ on $\R$. A primitive of $a(x) = \cos x$ is $A(x) = \sin x$, so the solutions are

Two readings. Every solution is periodic (period $2\pi$) and never vanishes unless $\lambda = 0$ — the sign of $\lambda$ is the sign of $y$ forever, since an exponential cannot cross zero. And the solution through $y(0) = y_0$ is $y_0\eu^{-\sin x}$: exactly one curve of the family through each initial point, the one-dimensional picture of Theorem 5.4 (2).

Example 5.6 (Guessing beats integrating)

Solve $y' + 2x\,y = x$ on $\R$. Variation of constants works ($A = x^2$, $\mu' = x\,\eu^{x^2}$, $\mu = \frac12\eu^{x^2} + \lambda$), but observing that the constant $y_p = \frac12$ solves the equation ($0 + 2x\cdot\frac12 = x$) is faster. With the homogeneous solutions $\lambda\,\eu^{-x^2}$:

Every solution converges to $\frac12$ extremely fast as $x \to \pm\infty$: the constant particular solution is an equilibrium that all solutions join. The insight: before launching the general method, spend ten seconds looking for an obvious particular solution (constant, monomial, multiple of the right-hand side); the structure theorem then finishes the job.

Example 5.8 (A complex right-hand side, two real answers)

Solve $y' - y = \cos x$ and $y' - y = \sin x$ in one stroke. Work in $\C$ with the right-hand side $\eu^{\iu x}$: trying $y_p = c\,\eu^{\iu x}$ gives $c(\iu - 1)\eu^{\iu x} = \eu^{\iu x}$, so

Since the equation has real coefficients, real and imaginary parts split: $\frac{\sin x - \cos x}2$ solves $y' - y = \cos x$, and $-\frac{\sin x + \cos x}2$ solves $y' - y = \sin x$ (check the first: derivative $\frac{\cos x + \sin x}2$, minus the function, gives $\cos x$). One complex line replaced two runs of variation of constants — the same economy that Method 5.13 systematizes for second order, and a recurring dividend of Chapter 3.