Undetermined Coefficients — Question 8

Question 8 – Solution

Strategy. Expand the trigonometric product before choosing resonance factors.

Step 1: Expose both forcing frequencies. The identity $\cos^3x=(3\cos x+\cos 3x)/4$ reveals simple resonance at both $\pm i$ and $\pm 3i$. A full trial is $$x(A\cos x+B\sin x+C\cos 3x+D\sin 3x).$$ There is no nonlinear dependence on the unknown $y$; the right-hand side is a known function of $x$.

Step 2: Match each harmonic. For $L=(D^2+1)(D^2+9)$, differentiation gives $L(x\sin x)=16\cos x$ and $L(x\sin 3x)=-48\cos 3x$. The cosine-weighted partners generate sine forcing and therefore have zero coefficients. Thus $$y_p=\frac{3x\sin x}{64}-\frac{x\sin 3x}{192},\qquad y_h=a\cos x+b\sin x+c\cos 3x+d\sin 3x.$$ The residual is $3\cos x/4+\cos 3x/4=\cos^3x$.

Step 3: Correct the initial curvature. The particular term is even, has value zero, and has $y_p''(0)=3/32-1/32=1/16$. The zero data require $a+c=0$, $-a-9c=-1/16$, and $b+3d=-b-27d=0$. Hence $$\boxed{y=\frac{x(9\sin x-\sin 3x)}{192}+\frac{\cos 3x-\cos x}{128}.}$$ Here $a=-1/128,c=1/128,b=d=0$. The cosine correction has second derivative $-1/16$ at zero and cancels the particular curvature; evenness verifies the odd initial derivatives.

Step 4: Prove growth and a global envelope. At $x_n=\pi/2+2\pi n$, the sine values are $1,-1$ and both cosines vanish, so $\boxed{y(x_n)=5x_n/96\to\infty}$. Also $9\sin x-\sin 3x=6\sin x+4\sin^3x$, whose absolute value is at most $10$. Therefore $$\boxed{|y(x)|\le\frac{5x}{96}+\frac 1{64}\quad(x\ge 0).}$$ The displayed envelopes are bounds, not assertions that equality holds at every point.