Nonconstant Coefficient IVP’s — Question 3

Question 3 – Solution

Strategy. Check the actual time solution first. Necessary transform limits can hold for functions of $s$ that are not transforms of the IVP solution.

Step 1: Establish nonexistence of an ordinary transform. Separation and the initial datum give $\boxed{y=e^{t^2}}$, which is smooth and global. For any fixed real $s$, however, $t^2-st\ge t^2/2$ for sufficiently large $t$. Thus $$\int_0^\infty e^{-st}y(t)\,dt=\int_0^\infty e^{t^2-st}\,dt=\infty.$$ The exact real transform domain is empty.

Step 2: Solve the formal transformed equation. Pretending the rules apply gives $sY-1+2Y'=0$, so $$Y'+\tfrac s2Y=\tfrac 12,\qquad \boxed{Y_C(s)=e^{-s^2/4}[C+\tfrac 12\int_0^s e^{u^2/4}\,du].}$$ This is a valid family of solutions to the differential equation in $s$. It is not yet a family of Laplace transforms.

Step 3: Check the deceptively correct limits. As $s\to\infty$, l’Hopital’s rule gives $$\frac{\int_0^s e^{u^2/4}\,du}{2e^{s^2/4}/s}\longrightarrow 1,$$ because the derivative ratio is $1/(1-2/s^2)\to 1$. Hence the integral contribution to $Y_C$ is asymptotic to $1/s$, while $Ce^{-s^2/4}$ decays faster. Every real $C$ therefore satisfies $$\boxed{Y_C(s)\to 0,\qquad sY_C(s)\to 1.}$$ These two necessary-looking conditions do not even select a unique formal branch here.

Step 4: Identify the invalid step. The ordinary integrals defining $Y$, the transform of $ty$, and the derivative transform do not converge for any real $s$. The time solution is not of exponential order, and direct growth proves failure, not merely absence of a sufficient hypothesis. Applying transform identities to these divergent integrals was unjustified. The limits in Step3 cannot repair that step. In contrast, changing the sign to $y'+2ty=0$ gives $e^{-t^2}$, whose transform exists for every real $s$ and can be verified directly. A differential equation in transform space is useful only after its relationship to convergent time integrals is established.