Nonhomogeneous Equations & Resonance

Study Sheet

Nonhomogeneous Equations & Resonance

Undetermined coefficients, and what forcing does

Undetermined Coefficients

Tip
Guess the shape of the forcing

For y′′+by′+cy=f(x)y'' + by' + cy = f(x): general solution == homogeneous ++ ONE particular solution. Guess ypy_p shaped like ff: polynomials for polynomials (4x4x gets ax+bax + b), AekxAe^{kx} for exponentials, Acos⁡+Bsin⁡A\cos + B\sin for sinusoids — substitute and match coefficients.

Tip
Resonance: multiply by x

When the forcing already SOLVES the homogeneous equation, the naive guess collapses — multiply it by xx. For y′′+4y=cos⁡2xy'' + 4y = \cos 2x: yp=x4sin⁡2xy_p = \tfrac{x}{4}\sin 2x, whose amplitude GROWS without bound. This is resonance — the mathematics behind pushed swings and shattered bridges.

Example
A full match, worked

y′′+3y′+2y=4xy'' + 3y' + 2y = 4x with yp=ax+by_p = ax + b: substituting gives 3a+2(ax+b)=4x3a + 2(ax + b) = 4x, so a=2a = 2 and b=−3b = -3: yp=2x−3y_p = 2x - 3. Matching term-by-term is bookkeeping, not cleverness — write the shape, then compare.

Side note
Significance: resonance is not a metaphor

The Broughton suspension bridge (1831) collapsed under soldiers marching in step; the Millennium Bridge in London wobbled dangerously on opening day from synchronized footfalls — both are the xsin⁡ωxx\sin\omega x growth of this topic in steel. Every tuned structure is designed to keep its natural frequencies away from expected forcing.

Try it
Try it: spot resonance before solving

Which forcing causes resonance in y′′+9y=f(x)y'' + 9y = f(x): cos⁡2x\cos 2x, cos⁡3x\cos 3x, or e3xe^{3x}? Work: homogeneous solutions oscillate at frequency 33, so cos⁡3x\cos 3x resonates (guess needs the extra xx); cos⁡2x\cos 2x and e3xe^{3x} do not — e3xe^{3x} is not a homogeneous solution here since roots are ±3i\pm3i, not 33.

Proofs & Why It Matters

Tip
Proof: general = homogeneous + particular

Let ypy_p solve Ly=fLy = f (writing LL for the whole left side). If yy is ANY solution of Ly=fLy = f, then L(y−yp)=f−f=0L(y - y_p) = f - f = 0 by linearity, so y−ypy - y_p is a homogeneous solution: y=yh+ypy = y_h + y_p. Conversely every yh+ypy_h + y_p solves Ly=fLy = f.

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

Tip
Proof: the resonant particular solution

For y′′+ω2y=cos⁡ωxy'' + \omega^2y = \cos\omega x, try y=kxsin⁡ωxy = kx\sin\omega x: y′=ksin⁡ωx+kωxcos⁡ωxy' = k\sin\omega x + k\omega x\cos\omega x and y′′=2kωcos⁡ωx−kω2xsin⁡ωxy'' = 2k\omega\cos\omega x - k\omega^2x\sin\omega x. Then y′′+ω2y=2kωcos⁡ωxy'' + \omega^2y = 2k\omega\cos\omega x — the xx-terms cancel — so k=12ωk = \tfrac{1}{2\omega} (=14= \tfrac14 when ω=2\omega = 2).

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

Going Deeper: Explanations & Worked Problems

Concept
The guessing table, and the logic behind it

Undetermined coefficients works because differentiation maps each function family into itself: polynomials to polynomials, ekxe^{kx} to itself, {cos⁡ωx,sin⁡ωx}\{\cos\omega x, \sin\omega x\} into each other.

So guess a general member of the forcing's family: for 4x4x, guess ax+bax + b (BOTH terms — the derivative of axax makes constants, so a bare axax cannot balance); for e2xe^{2x}, guess Ae2xAe^{2x}; for cos⁡2x\cos 2x, guess Acos⁡2x+Bsin⁡2xA\cos2x + B\sin2x (both, since y′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 xx (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.

Example
Worked: polynomial forcing, all coefficients matched

Find the general solution of y′′+3y′+2y=4xy'' + 3y' + 2y = 4x.

Step 1 — homogeneous part: r2+3r+2=(r+1)(r+2)=0r^2 + 3r + 2 = (r+1)(r+2) = 0, so yh=c1e−x+c2e−2xy_h = c_1e^{-x} + c_2e^{-2x}.

Step 2 — guess yp=ax+by_p = ax + b (not a homogeneous solution — no xx-multiplication needed): yp′=ay_p' = a, yp′′=0y_p'' = 0.

Step 3 — substitute: 0+3a+2(ax+b)=4x0 + 3a + 2(ax + b) = 4x, i.e. 2a x+(3a+2b)=4x+02a\,x + (3a + 2b) = 4x + 0.

Step 4 — match coefficient BY coefficient: xx: 2a=4⇒a=22a = 4 \Rightarrow a = 2; constants: 3(2)+2b=0⇒b=−33(2) + 2b = 0 \Rightarrow b = -3.

Step 5 — assemble: y=c1e−x+c2e−2x+2x−3y = c_1e^{-x} + c_2e^{-2x} + 2x - 3.

Step 6 — long-run reading: both homogeneous pieces decay, so EVERY solution approaches the line 2x−32x - 3 — the particular solution is the system's steady response, the homogeneous part its fading memory of initial conditions.

Example
Worked: resonance versus near-resonance, side by side

Compare y′′+4y=cos⁡3xy'' + 4y = \cos 3x against y′′+4y=cos⁡2xy'' + 4y = \cos 2x (natural frequency 22). NON-RESONANT: guess yp=Acos⁡3xy_p = A\cos3x; then yp′′=−9Acos⁡3xy_p'' = -9A\cos3x, so (−9A+4A)cos⁡3x=cos⁡3x(-9A + 4A)\cos3x = \cos3x gives A=−15A = -\tfrac15 — a bounded steady oscillation, done.

RESONANT: the naive guess Acos⁡2xA\cos2x gives (−4A+4A)cos⁡2x=0≠cos⁡2x(-4A + 4A)\cos2x = 0 \ne \cos2x — unsolvable, exactly because cos⁡2x\cos2x solves the homogeneous equation. Multiply by xx: yp=x(Acos⁡2x+Bsin⁡2x)y_p = x(A\cos2x + B\sin2x). Differentiate twice (product rule, carefully): yp′′=−4x(Acos⁡2x+Bsin⁡2x)+2(−2Asin⁡2x+2Bcos⁡2x)y_p'' = -4x(A\cos2x + B\sin2x) + 2(-2A\sin2x + 2B\cos2x); adding 4yp4y_p cancels the xx-terms, leaving 4Bcos⁡2x−4Asin⁡2x=cos⁡2x4B\cos2x - 4A\sin2x = \cos2x: B=14B = \tfrac14, A=0A = 0, so yp=x4sin⁡2xy_p = \tfrac x4\sin2x. The amplitude GROWS linearly forever — same equation shape, one frequency changed, qualitatively different fate.