Linear Equations — Question 10

Question 10 – Solution

Strategy. Keep the parameter visible during integration, and isolate the exceptional exponent before dividing by $a-1$.

Step 1: Integrate with the initial value. With $\mu=e^{ax}$, $$(e^{ax}y)'=e^{(a-1)x},\qquad e^{ax}y=1+\int_0^x e^{(a-1)t}\,dt.$$ Consequently $$\boxed{y(x)=\begin{cases} \displaystyle\frac{e^{-x}+(a-2)e^{-ax}}{a-1},&a\ne 1,\\[6pt] (1+x)e^{-x},&a=1. \end{cases}}$$ Each formula is defined on $\mathbb R$.

Step 2: Verify. At zero the numerator for $a\ne 1$ is $a-1$, so $y(0)=1$; the exceptional formula also gives $1$. Differentiating the definite-integral identity above gives $y'+ay=e^{-x}$. At $a=1$, a direct check is $y'=-xe^{-x}$ and $y'+y=e^{-x}$.

Step 3: Remove the apparent parameter singularity. For $a\ne 1$, rewrite the result as $$y=e^{-ax}\left[1+\frac{e^{(a-1)x}-1}{a-1}\right].$$ For fixed $x$, the quotient tends to $x$ as $a\to 1$, by the derivative of the exponential (also at $x=0$). Hence $y\to(1+x)e^{-x}$. The singularity in the divided parameter formula is removable.

Step 4: Long-time behavior. If $a>0$ and $a\ne 1$, both exponentials decay, so $y\to 0$; the same holds at $a=1$. If $a=0$, $y=2-e^{-x}\to 2$. If $a<0$, the coefficient $(a-2)/(a-1)$ is positive and $e^{-ax}$ grows without bound, so $y\to+\infty$. Thus $$\boxed{y\text{ is bounded on }[0,\infty)\quad\Longleftrightarrow\quad a\ge 0.}$$