Nonconstant Coefficient IVP's — Question 9

Question 9

Use ordinary one-sided Laplace integrals for real $s$. Where justified, write $Y(s)=\mathcal L\{y\}(s)$ and use $\mathcal L\{ty\}=-Y'(s)$. Check existence and initial compatibility before treating a formal solution in $s$ as a transform.

An unknown parameter $\alpha>0$ appears in $$(1+t)y'+\alpha y=0,\qquad y(0)=1.$$ An exact measurement gives $\int_0^\infty y(t)\,dt=1/2$. A second proposed measurement is $\int_0^\infty ty(t)\,dt=1/3$.

Tasks

1. Derive the differential equation for $Y$ and solve it as a convergent parameter integral for $s>0$.

2. Identify and verify the time solution for general $\alpha>0$.

3. Determine which parameters admit a finite area, then recover $\alpha$ from the first measurement.

4. Test the proposed first moment for consistency, and state the exact real transform domain of the recovered solution.

Question 9 – Solution

Strategy. Solve the parameter family before fitting measurements. A boundary transform value is meaningful only when the actual time integral converges.

Step 1: Derive and solve the transform equation. Using $\mathcal L\{y'\}=sY-1$ and $\mathcal L\{ty'\}=-Y-sY'$ gives $$-sY'+(s+\alpha-1)Y=1.$$ The integrating factor for the divided equation is $e^{-s}s^{1-\alpha}$, so its product with $Y$ has derivative $-e^{-s}s^{-\alpha}$. The transform bound at infinity selects $$\boxed{Y(s)=e^s s^{\alpha-1}\int_s^\infty e^{-u}u^{-\alpha}\,du,\qquad s>0.}$$ The alternative homogeneous term $Ce^s s^{\alpha-1}$ violates decay as $s\to\infty$.

Step 2: Verify the parameter family in time. Separation gives $\boxed{y(t)=(1+t)^{-\alpha}}$. Its derivative is $-\alpha(1+t)^{-\alpha-1}$, so the equation and initial value hold. The coefficient $1+t$ is nonzero on $t\ge 0$, establishing regular-IVP uniqueness. Substitution $u=s(1+t)$ in the forward integral gives exactly the parameter expression above, verifying its inversion.

Step 3: Use the convergent area to identify the parameter. The positive power tail has finite area exactly when $\alpha>1$, and then $$\int_0^\infty(1+t)^{-\alpha}\,dt=\frac 1{\alpha-1}.$$ Equating this to $1/2$ gives the unique value $\boxed{\alpha=3}$. This uses convergence of the integral, not a formal substitution into an expression initially derived only for $s>0$.

Step 4: Test the second datum rather than refit. A finite first moment requires $\alpha>2$. Substituting $x=1+t$ gives $$\int_0^\infty\frac{t}{(1+t)^\alpha}\,dt =\frac 1{\alpha-2}-\frac 1{\alpha-1} =\frac 1{(\alpha-1)(\alpha-2)}.$$ At $\alpha=3$, it is $\boxed{1/2}$, not $1/3$, so the proposed data are inconsistent with this family. The recovered positive tail is $(1+t)^{-3}$; its exact transform domain is $\boxed{[0,\infty)}$. Positive $s$ gives exponential damping, zero gives the finite area $1/2$, and negative $s$ makes the weighted power tail diverge.