Nonconstant Coefficient IVP’s — Question 8

Question 8 – Solution

Strategy. Singular coefficients can erase initial data from the transformed equation. Recover admissibility and the remaining amplitude separately.

Step 1: Check compatibility and time powers. At zero, the equation requires $\boxed{a=0}$. For $t>0$, trial powers give $m(m-1)+2m-2=(m-1)(m+2)=0$. The two independent solutions are $t$ and $t^{-2}$, so $y=Ct+Dt^{-2}$. They span the solutions on $(0,\infty)$ because the normalized equation is regular there. Continuity at zero forces $D=0$; the initial slope then sets $C=b$.

Step 2: Transform the polynomial coefficients. For sufficiently large $s$, $$\mathcal L\{t^2y''\}=\frac{d^2}{ds^2}[s^2Y-sa-b] =2Y+4sY'+s^2Y'',$$ while $2\mathcal L\{ty'\}=-2Y-2sY'$. Adding $-2Y$ gives $$\boxed{s^2Y''+2sY'-2Y=0.}$$ Both $a$ and $b$ vanished under differentiation; their disappearance is not permission to ignore the time-domain data.

Step 3: Select the transform and its amplitude. Trial powers of $s$ have exponents one and $-2$, yielding $Y=As+B/s^2$ on $s>0$. A genuine transform of an exponential-order continuous function tends to zero at large $s$, so $A=0$. This still leaves $B$ arbitrary. The supplied initial-slope limit gives $\boxed{B=b}$ and hence $Y=b/s^2$. The initial-value limit $sY\to 0$ also agrees with the necessary datum $a=0$.

Step 4: Verify and classify uniqueness. All and only the data $\boxed{a=0,\ b\in\mathbb R}$ are admissible, with unique solution $\boxed{y=bt}$. Substitution gives $0+2tb-2bt=0$, and both data hold. For $b\ne 0$, the exact domain is $s>0$; the signed linear tail diverges at or below zero. For $b=0$, the zero function has transform zero for every real $s$. Specifying only $y(0)=0$ leaves the entire family $bt$, so it does not determine a unique solution at this singular point.