Undetermined Coefficients — Question 8

Question 8 – Solution

Strategy. The standard method depends on closure in a finite differentiation family; a plausible-looking trial outside that family need not close.

Step 1: Classify the forcings. The first two belong to the standard families. The third does too after $\cos^2t=(1+\cos 2t)/2$. The functions $e^{t^2}$ and $1/(1+t)$ do not: successive derivatives introduce polynomial factors of ever-increasing degree in the first case and poles of ever-increasing order in the second. The usual finite constant-coefficient matching prescription does not apply to them.

Step 2: Test one inverse power. Put $x=1+t>0$. For $y_p=A/x$, $$y_p''+y_p=A/x+2A/x^3.$$ Equality to $1/x$ would require $A(x^2+2)=x^2$ for all $x>0$. Matching $x^2$ forces $A=1$, while matching the constant term forces $2A=0$, a contradiction.

Step 3: Exclude any finite inverse-power sum. If such a sum were nonzero, choose its largest index $N$ with $a_N\ne 0$. Its second derivative contains $$a_NN(N+1)x^{-N-2}.$$ No other differentiated term, undifferentiated term or forcing term has that most singular power. Equivalently, multiply the proposed equation by $x^{N+2}$ and let $x\downarrow 0$: the left side tends to $a_NN(N+1)\ne 0$, while the right side tends to zero. The zero sum also fails to give the nonzero forcing.

Step 4: Separate method failure from nonexistence. On $I$, the normalized coefficients and the forcing $1/(1+t)$ are continuous. For every $t_0\in I$ and every finite value and slope at $t_0$, the regular linear theorem gives exactly one solution on $I$. A failed finite ansatz is a limitation of the trial class, not a proof that the differential equation lacks solutions.