Starting from the definition of stability, examine the…

Starting from the definition of stability, examine the stability of the solution of the differential equation $$\frac{dy}{dx}=-a^2 y,\quad a\neq0,\quad y(x_0)=y_0.$$

Step 1 · Explicit solution

The equation is a first-order linear homogeneous ODE with constant coefficient. Separating variables: $$\frac{dy}{y}=-a^2\,dx \implies \ln|y|=-a^2 x+C \implies y(x)=y_0\,e^{-a^2(x-x_0)}.$$ Thus, for the initial condition $y(x_0)=y_0$, the solution is $$\varphi(x;\,x_0,y_0)=y_0\,e^{-a^2(x-x_0)}.$$

Step 2 · Definitions of stability

Let $\varphi(x)=\varphi(x;x_0,y_0)$ be a given solution. The solution is called:

• stable (in the sense of Lyapunov) if for every $\varepsilon>0$ there exists $\delta=\delta(\varepsilon,x_0)>0$ such that $|\tilde{y}_0-y_0|<\delta$ implies $$|\varphi(x;x_0,\tilde{y}_0)-\varphi(x;x_0,y_0)|<\varepsilon \quad\text{for all }x\geq x_0;$$
• asymptotically stable if it is stable and additionally $$\lim_{x\to+\infty}|\varphi(x;x_0,\tilde{y}_0)-\varphi(x;x_0,y_0)|=0;$$
• unstable if it is not stable.

Step 3 · Perturbation of the initial condition

Let $\tilde{y}_0=y_0+\eta$ be a perturbed initial condition, where $\eta$ is a small real number. The corresponding solution is: $$\varphi(x;x_0,\tilde{y}_0)=(y_0+\eta)\,e^{-a^2(x-x_0)}.$$ The difference between the perturbed and the original solution is: $$\Delta(x)=\varphi(x;x_0,\tilde{y}_0)-\varphi(x;x_0,y_0) =\eta\,e^{-a^2(x-x_0)}.$$

Step 4 · Estimation of the difference

For all $x\geq x_0$, since $a^2>0$: $$e^{-a^2(x-x_0)}\leq 1,$$ so $$|\Delta(x)|=|\eta|\,e^{-a^2(x-x_0)}\leq|\eta| \quad\text{for all }x\geq x_0.$$

Step 5 · Verification of Lyapunov stability

Given any $\varepsilon>0$, choose $\delta=\varepsilon$. Then: $$|\tilde{y}_0-y_0|=|\eta|<\delta=\varepsilon \implies |\Delta(x)|=|\eta|\,e^{-a^2(x-x_0)}\leq|\eta|<\varepsilon \quad\text{for all }x\geq x_0.$$ The definition of Lyapunov stability is satisfied. Hence the solution is stable.

Step 6 · Verification of asymptotic stability

Since $a^2>0$: $$\lim_{x\to+\infty}|\Delta(x)| =\lim_{x\to+\infty}|\eta|\,e^{-a^2(x-x_0)}=0,$$ regardless of the (sufficiently small) perturbation $\eta$. The difference between any perturbed solution and the original solution tends to zero as $x\to+\infty$. Hence the solution is asymptotically stable.

Step 7 · Geometric interpretation

Every solution $y(x)=y_0\,e^{-a^2(x-x_0)}$ decays exponentially to zero as $x\to+\infty$, regardless of the initial value $y_0$, since $a^2>0$ for all $a\neq0$. All trajectories are therefore attracted toward the zero solution: nearby solutions not only stay close but actually converge to the same limit. This is the hallmark of asymptotic stability.

Step 8 · Conclusion

$$\boxed{ \text{The solution }y(x)=y_0\,e^{-a^2(x-x_0)} \text{ is \textbf{asymptotically stable} for all }a\neq0. }$$ Key observations:

• Since $a^2>0$ always (regardless of the sign of $a$), the exponent $-a^2(x-x_0)$ is always negative for $x>x_0$, guaranteeing exponential decay of all solutions.
• The stability is uniform in $x_0$: the choice $\delta=\varepsilon$ does not depend on $x_0$.
• Asymptotic stability follows directly from $e^{-a^2(x-x_0)}\to0$ as $x\to+\infty$.

Summary Table

Additional Notes

• Since $a^2>0$ always (regardless of the sign of $a$), the exponent $-a^2(x-x_0)$ is always negative for $x>x_0$, guaranteeing exponential decay of all solutions.
• The stability is uniform in $x_0$: the choice $\delta=\varepsilon$ does not depend on $x_0$.
• Asymptotic stability follows directly from $e^{-a^2(x-x_0)}\to0$ as $x\to+\infty$.