Undetermined Coefficients — Question 4

Question 4 – Solution

Strategy. Combine overlapping forcing families before choosing the trial, then apply the resonance correction separately to each distinct exponential.

Step 1: Simplify the forcing. The right side becomes $$(t+2)e^t+e^{-t}.$$ The characteristic roots of $y''-y=0$ are $1,-1$, both simple. Therefore each exponential family is resonant once.

Step 2: Choose a nonredundant trial. Use $$y_p=e^t(at^2+bt)+ct e^{-t}.$$ A separate trial for the $e^t$ inside $2\cosh t$ would repeat the $te^t$ term already in the first family’s trial. Combining the forcing first avoids two coefficients multiplying the same function.

Step 3: Match coefficients. Product differentiation gives $$(D^2-1)(e^tF)=e^t(F''+2F'),\qquad (D^2-1)(e^{-t}G)=e^{-t}(G''-2G').$$ Thus the trial produces $e^t(4at+2a+2b)-2ce^{-t}$. Matching gives $a=1/4$, $b=3/4$, $c=-1/2$, so $$\boxed{y_p=e^t(t^2/4+3t/4)-\tfrac 12te^{-t}.}$$

Step 4: Fit and verify. The particular data are $(0,1/4)$. With $y=y_p+Ce^t+De^{-t}$, the zero data require $C+D=0$, $C-D=-1/4$, hence $C=-1/8$, $D=1/8$. Therefore $$\boxed{y=e^t(t^2/4+3t/4-1/8)+e^{-t}(1/8-t/2).}$$ Its value is $-1/8+1/8=0$ and its slope is $1/4-1/8-1/8=0$. The two added exponentials are homogeneous, so the coefficient-matched forcing is preserved.