Nonconstant Coefficient IVP’s — Question 6

Question 6 – Solution

Strategy. Keep the initial derivative term from the unmultiplied second derivative. The coefficient $1+t$ never vanishes on the time domain.

Step 1: Transform with both data. The three contributions are $$\mathcal L\{ty''\}=-2sY-s^2Y',\quad \mathcal L\{y''\}=s^2Y-1,\quad \mathcal L\{y'\}=sY.$$ Hence $-s^2Y'+(s^2-s)Y=1$, or $$\boxed{Y'-(1-1/s)Y=-1/s^2},\qquad s>0.$$

Step 2: Select the parameter integral. The integrating factor is $se^{-s}$, so $(se^{-s}Y)'=-e^{-s}/s$. An exponential-order continuous time solution has $Y=O(1/s)$ for large $s$, and thus $se^{-s}Y\to 0$. Therefore $$\boxed{Y(s)=\frac{e^s}{s}\int_s^\infty\frac{e^{-u}}u\,du.}$$ An additional term $Ce^s/s$ violates the large-parameter bound unless $C=0$. Existence and the growth condition are verified by the recovered solution.

Step 3: Identify and check the inverse. In time, the left side is $[(1+t)y']'$. Integrating with the data gives $y'=1/(1+t)$ and $$\boxed{y(t)=\ln(1+t).}$$ It has $y(0)=0$, $y'(0)=1$, $y''=-1/(1+t)^2$, and zero equation residual. The coefficient is regular, so uniqueness applies. Integrating its transform by parts gives $$Y(s)=\frac 1s\int_0^\infty\frac{e^{-st}}{1+t}\,dt =\frac{e^s}{s}\int_s^\infty\frac{e^{-u}}u\,du,$$ where the upper boundary vanishes for $s>0$.

Step 4: Compare growth rates and domains. For $t>0$, $0<\ln(1+t)<t$. Multiplying by the positive kernel and integrating gives $$\boxed{0<Y(s)<\frac 1{s^2},\qquad s>0.}$$ The exact domain is $s>0$: logarithmic growth is dominated by the exponential kernel above zero, whereas its positive time integral diverges at zero and for negative parameters. Unboundedness by itself does not prevent a Laplace transform; growth relative to the kernel is what matters.