Undetermined Coefficients — Question 1

Question 1 – Solution

Strategy. Differentiation creates lower polynomial powers, so include them before matching every coefficient.

Step 1: Complete the trial. For $At^2$, the residual is $2At^2-6At+2A$, which cannot equal $t^2$. Use $y_p=at^2+bt+c$. The nonzero coefficient of $y$ preserves the highest polynomial degree, so degree two is necessary and sufficient here.

Step 2: Match coefficients. Substitution yields $$2at^2+(2b-6a)t+(2a-3b+2c)=t^2.$$ Thus $a=1/2$, $b=3/2$, $c=7/4$, giving $$\boxed{y_p=\tfrac 12t^2+\tfrac 32t+\tfrac 74.}$$ The displayed coefficient equations verify the residual identically.

Step 3: Fit the homogeneous correction. The characteristic roots are $1,2$, so $y=y_p+Ce^t+De^{2t}$. At zero, $$C+D=-7/4,\qquad C+2D=-3/2.$$ Hence $C=-2$, $D=1/4$, and $$\boxed{y=\tfrac 12t^2+\tfrac 32t+\tfrac 74-2e^t+\tfrac 14e^{2t}.}$$ Its value is $7/4-2+1/4=0$ and its slope is $3/2-2+1/2=0$. Adding homogeneous modes does not change the verified forcing.

Step 4: Prove polynomial uniqueness. The difference of two polynomial particular solutions would be a polynomial $h$ with $h''-3h'+2h=0$. If $h$ were nonzero with degree $m$ and leading coefficient $a_m$, the leading term of this residual would be $2a_mt^m$, which cannot vanish. Thus $h=0$. Nonpolynomial homogeneous additions still give infinitely many other particular solutions.