Nonhomogeneous Equations & Resonance

Why the table works: derivatives of polynomials are polynomials of no higher degree, derivatives of $e^{kx}$ are multiples of $e^{kx}$, and derivatives of $\cos\omega x$ and $\sin\omega x$ stay among $\cos\omega x, \sin\omega x$. So the left side applied to the guess lands in the same family as $f$, and matching coefficients is a finite linear system. The homogeneous part is added because the left side sends it to zero. For $y'' + 3y' + 2y = 4x$ (worked below) that part decays, and every solution approaches $y_p = 2x - 3$.

Why multiplying by $x$ helps: if $g$ solves the homogeneous equation, then substituting $xg$ leaves $x$ times the homogeneous expression for $g$, which is zero, plus terms built from $g'$ that are not zero. For $y'' + 4y = \cos 2x$: $(x\sin 2x)'' + 4x\sin 2x = 4\cos 2x$, so $y_p = \tfrac x4\sin 2x$. This $y_p$ has $y_p(0) = y_p'(0) = 0$, so it is the motion from rest, and its swings grow between the lines $\pm\tfrac x4$.

Why $ax + b$ and not just $ax$: substituting $y_p = ax$ gives $3a + 2ax = 4x$, which needs $2a = 4$ and $3a = 0$ at once — impossible. Differentiating $ax$ produces the constant $3a$, and the constant $b$ in the guess is there to absorb it.

Each family is closed under differentiation, so the left side applied to the guess returns to the family, and comparing coefficients gives a square linear system. Both cosine and sine are needed because the $by'$ term turns one into the other. Adding particular solutions works because the left side $L$ is linear: $L[y_1 + y_2] = L[y_1] + L[y_2] = f_1 + f_2$.

Step 3 — substitute: $Ae^x - Ae^x - 2Ae^x = -2Ae^x = 6e^x$, so $A = -3$.

Step 4 — $y = c_1e^{2x} + c_2e^{-x} - 3e^x$. A shortcut worth knowing: for $f = e^{kx}$ with $k$ not a root, substituting $Ae^{kx}$ gives $A(k^2 + bk + c)e^{kx}$, so $A$ is the forcing's coefficient divided by the characteristic polynomial at $k$; here $\dfrac{6}{1 - 1 - 2} = -3$.

Step 3 — substitute and collect: the $\cos x$ terms give $-A + 2B + 5A = 4A + 2B$, the $\sin x$ terms $-B - 2A + 5B = 4B - 2A$. Matching $10\cos x + 0\sin x$: $4A + 2B = 10$ and $-2A + 4B = 0$, so $A = 2B$, $10B = 10$, $B = 1$ and $A = 2$.

Step 4 — the data: $y(0) = c_1 + 2 = 0$ gives $c_1 = -2$, and $y'(0) = -c_1 + 2c_2 + 1 = 0$ gives $c_2 = -\tfrac32$.

Step 5 — $y = e^{-x}\left(-2\cos 2x - \tfrac32\sin 2x\right) + 2\cos x + \sin x$.

Step 6 — read it: the transient dies like $e^{-x}$, leaving the steady state $2\cos x + \sin x = \sqrt5\cos(x - \varphi)$ with $\varphi = \arctan\tfrac12$ — amplitude $\sqrt5 \approx 2.24$, at the driving frequency $1$ rather than the natural frequency $2$.

Step 3 — $y_p'' + y_p = (2ax + 2a + 2b)e^x$ must equal $xe^x$, so $2a = 1$ and $2a + 2b = 0$: $a = \tfrac12$, $b = -\tfrac12$.

Step 4 — $y_p = \tfrac12(x - 1)e^x$. The constant $b$ was necessary: with $y_p = axe^x$ alone, the constant terms would demand $2a = 0$, contradicting $2a = 1$.

Why the multiplicity: when $k$ is a root of multiplicity $m$, the functions $e^{kx}, xe^{kx}, \ldots, x^{m-1}e^{kx}$ all solve the homogeneous equation, so the left side sends each of them to zero and the first power of $x$ that can balance the forcing is $x^m$. Beats are the near-miss version of resonance: as $\omega \to \omega_0$ the slow envelope $\sin\frac{(\omega_0 - \omega)x}{2}$ stretches out, and over any fixed stretch of $x$ the response grows almost linearly, approaching the resonant $x\sin\omega_0x$.

Step 3 — $y_p'' - y_p = 2Ae^x = e^x$, so $A = \tfrac12$ and $y_p = \tfrac12xe^x$. The $x$-terms cancelled, as they must: the part of $Axe^x$ that looks like a homogeneous solution contributes nothing, and what is left is a constant multiple of $e^x$.

Step 3 — substitute: $Ae^{2x}\left[(2 + 8x + 4x^2) - 4(2x + 2x^2) + 4x^2\right] = 2Ae^{2x}$, so $A = \tfrac12$ and $y_p = \tfrac12x^2e^{2x}$.

Step 4 — the general solution is $y = (c_1 + c_2x)e^{2x} + \tfrac12x^2e^{2x}$.

Step 3 — the general solution is $y = c_1\cos 5x + c_2\sin 5x + \cos 4x$; $y(0) = c_1 + 1 = 0$ gives $c_1 = -1$, and $y'(0) = 5c_2 = 0$ gives $c_2 = 0$.

Step 4 — $y = \cos 4x - \cos 5x = 2\sin\tfrac x2\,\sin\tfrac{9x}{2}$.

Step 5 — read it: a fast oscillation at frequency $\tfrac92$ inside the slow envelope $\pm2\sin\tfrac x2$, which swells to $2$ at $x = \pi$ and shrinks back to $0$ at $x = 2\pi$. The system alternately absorbs energy from the forcing and hands it back.

One particular solution plus the whole homogeneous family is therefore the complete solution set. $\blacksquare$

The linear-in-$x$ amplitude is forced: bounded guesses die because they already solve the homogeneous equation. $\blacksquare$

So guess a general member of the forcing's family: for $4x$, guess $ax + b$ (both terms — the derivative of $ax$ makes constants, so a bare $ax$ cannot balance); for $e^{2x}$, guess $Ae^{2x}$; for $\cos 2x$, guess $A\cos2x + B\sin2x$ (both, since $y'$ turns cosine into sine); for products, multiply the guesses. The exception: if the guess already solves the homogeneous equation, substituting it gives zero on the left and the coefficient equations become unsolvable — multiply the guess by $x$ (twice for a double root). That is resonance, and physically it is why forcing a system at its natural frequency produces growing oscillations rather than a steady response.

Step 1 — homogeneous part: $r^2 + 3r + 2 = (r+1)(r+2) = 0$, so $y_h = c_1e^{-x} + c_2e^{-2x}$.

Step 2 — guess $y_p = ax + b$ (not a homogeneous solution — no $x$-multiplication needed): $y_p' = a$, $y_p'' = 0$.

Step 3 — substitute: $0 + 3a + 2(ax + b) = 4x$, i.e. $2a\,x + (3a + 2b) = 4x + 0$.

Step 4 — match coefficient by coefficient: $x$: $2a = 4 \Rightarrow a = 2$; constants: $3(2) + 2b = 0 \Rightarrow b = -3$.

Step 5 — assemble: $y = c_1e^{-x} + c_2e^{-2x} + 2x - 3$.

Step 6 — long-run reading: both homogeneous pieces decay, so every solution approaches the line $2x - 3$ — the particular solution is the system's steady response, the homogeneous part its fading memory of initial conditions.

Resonant: the naive guess $A\cos2x$ gives $(-4A + 4A)\cos2x = 0 \ne \cos2x$ — unsolvable, exactly because $\cos2x$ solves the homogeneous equation. Multiply by $x$: $y_p = x(A\cos2x + B\sin2x)$. Differentiate twice (product rule, carefully): $y_p'' = -4x(A\cos2x + B\sin2x) + 2(-2A\sin2x + 2B\cos2x)$; adding $4y_p$ cancels the $x$-terms, leaving $4B\cos2x - 4A\sin2x = \cos2x$: $B = \tfrac14$, $A = 0$, so $y_p = \tfrac x4\sin2x$. The amplitude grows linearly forever — same equation shape, one frequency changed, qualitatively different fate.