Laplace Transforms — Question 6

Question 6 – Solution

Strategy. Use the singular-endpoint constraint before transforming; the transform’s boundary condition replaces its integration constant.

Step 1: Enforce compatibility. At $t=0$, continuity of the equation gives $3y''(0)+1=0$, so $\boxed{y''(0)=-1/3}$. The leading coefficient $t$ vanishes at the origin; normalizing by it would produce singular coefficients there.

Step 2: Transform carefully. Writing $c=y''(0)$, the four transforms are $$-\frac{d}{ds}(s^3Y-s^2-c),\quad 3(s^2Y-s),\quad -\frac{d}{ds}(sY-1),\quad Y.$$ Their sum is $-(s^3+s)Y'-s=0$. Thus $Y'=-1/(1+s^2)$ for $s>0$. Because the transform of a continuous exponential-order function tends to zero as $s\to\infty$, the constant is fixed: $$\boxed{Y(s)=\int_s^\infty\frac{dv}{1+v^2}=\arctan(1/s),\qquad s>0.}$$

Step 3: Invert an integral representation. Since $Y(s)=\int_0^1 s/(s^2+u^2)\,du$, absolute convergence for $\operatorname{Re}s>0$ permits interchange of integrals. Therefore $$\boxed{y(t)=\int_0^1\cos(ut)\,du=\frac{\sin t}{t}\ (t>0),\qquad y(0)=1.}$$ For $t>0$, this function satisfies $t y''+2y'+t y=0$. Differentiating gives the required third-order equation; its analytic extension satisfies the equation at zero too.

Step 4: Prove analytic uniqueness. For $y=\sum_{n\geq 0}c_nt^n$, coefficient matching gives $$(n+1)\bigl((n+2)(n+3)c_{n+2}+c_n\bigr)=0\quad(n\geq 0).$$ The seeds $c_0=1,c_1=0$ uniquely give $c_{2k}=(-1)^k/(2k+1)!$ and $c_{2k+1}=0$, an everywhere convergent series. Ordinary uniqueness on $t>0$ extends local uniqueness to the whole positive axis. Finally $Y=s^{-1}-(3s^3)^{-1}+(5s^5)^{-1}+O(s^{-7})$, agreeing with $y(0)=1$, $y''(0)=-1/3$, and $y^{(4)}(0)=1/5$ in $Y\sim\sum y^{(j)}(0)/s^{j+1}$.