Nonconstant Coefficient IVP’s — Question 1

Question 1 – Solution

Strategy. Multiplication by time makes a differential equation in the transform variable. Its integration constant must be selected using transform behavior.

Step 1: Differentiate the full derivative transform. Here $\mathcal L\{y'\}=sY$, so $\mathcal L\{ty'\}=-(Y+sY')$. The transformed equation is $$sY-Y-sY'+Y=1/s,\qquad \boxed{Y'-Y=-1/s^2},\quad s>0.$$ Treating $\mathcal L\{ty'\}$ as only $-sY'$ would introduce an erroneous extra term.

Step 2: Select the integration constant. Multiply by $e^{-s}$. A bounded function has $Y=O(1/s)$ as $s\to\infty$, hence $e^{-s}Y\to 0$. Integration from $s$ to infinity gives $$\boxed{Y(s)=e^s\int_s^\infty\frac{e^{-u}}{u^2}\,du.}$$ The other homogeneous term is $Ce^s$, incompatible with that bound unless $C=0$. The inverse identified below is bounded, so this selection is self-consistent and verified rather than assumed without a check.

Step 3: Identify and forward-check the inverse. The original left side is $[(1+t)y]'$, giving $(1+t)y=t$ and $$\boxed{y(t)=\frac{t}{1+t}.}$$ Its derivative is $(1+t)^{-2}$, so $(1+t)y'+y=1$ and $y(0)=0$. For $s>0$, substitution $u=s(1+t)$ gives $\mathcal L\{(1+t)^{-1}\}=e^s\int_s^\infty e^{-u}/u\,du$. Thus the forward transform is $1/s-e^s\int_s^\infty e^{-u}/u\,du$, equal to the boxed parameter integral by integration by parts. The regular first-order IVP is unique.

Step 4: Check domain and onset. Since $y\to 1$ and is nonnegative, its exact domain is $\boxed{s>0}$. At the origin $y'(0)=1$, and the substitution $v=st$ gives $$s^2Y(s)=\int_0^\infty e^{-v}\frac{v}{1+v/s}\,dv\longrightarrow 1.$$ The integrand is dominated by $ve^{-v}$, so dominated convergence justifies the limit. Thus $Y\sim 1/s^2$, matching the zero initial value and unit initial slope.