Linear Equations — Question 4

Question 4 – Solution

Strategy. Separate the forced response from the homogeneous term. A condition at infinity can select the constant just as an initial value can.

Step 1: General solution. With $\mu=e^{-x}$, $$(e^{-x}y)'=e^{-2x},\qquad e^{-x}y=-\frac 12e^{-2x}+C.$$ Thus $\boxed{y=Ce^x-\tfrac 12e^{-x}}$; each formula is defined on $\mathbb R$.

Step 2: Impose boundedness. As $x\to\infty$, the second term tends to zero. If $C\ne 0$, the term $Ce^x$ is unbounded and cannot be canceled by that decaying term. Therefore $$\boxed{y_b=-\frac 12e^{-x},\qquad y_b(0)=-\frac 12.}$$ This solution is bounded and tends to zero, proving both existence and uniqueness under the stated boundedness condition.

Step 3: Perturb the selected initial value. Since $y(0)=C-1/2$, changing it to $-1/2+\varepsilon$ gives $C=\varepsilon$. Consequently $$\boxed{y_\varepsilon-y_b=\varepsilon e^x.}$$ For $\varepsilon>0$ the perturbed solution tends to $+\infty$; for $\varepsilon<0$ it tends to $-\infty$. Direct differentiation verifies $y'-y=e^{-x}$ for every $C$. The vanishing forcing does not remove a growing homogeneous component.