Starting from the definition of stability, examine the…

Starting from the definition of stability, examine the stability of the solution of the differential equation $$\frac{dy}{dt}=a^2 y,\quad a\neq0,\quad y(t_0)=y_0.$$

Step 1 · Explicit solution

The equation is a first-order linear homogeneous ODE with constant coefficient $a^2>0$. Separating variables: $$\frac{dy}{y}=a^2\,dt \implies \ln|y|=a^2 t+C \implies y(t)=y_0\,e^{a^2(t-t_0)}.$$ Thus, for the initial condition $y(t_0)=y_0$, the solution is $$\varphi(t;\,t_0,y_0)=y_0\,e^{a^2(t-t_0)}.$$

Step 2 · Definitions of stability

Let $\varphi(t)=\varphi(t;t_0,y_0)$ be a given solution. The solution is called:

• stable (in the sense of Lyapunov) if for every $\varepsilon>0$ there exists $\delta=\delta(\varepsilon,t_0)>0$ such that $|\tilde{y}_0-y_0|<\delta$ implies $$|\varphi(t;t_0,\tilde{y}_0)-\varphi(t;t_0,y_0)|<\varepsilon \quad\text{for all }t\geq t_0;$$
• asymptotically stable if it is stable and additionally $$\lim_{t\to+\infty}|\varphi(t;t_0,\tilde{y}_0)-\varphi(t;t_0,y_0)|=0;$$
• 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 ($\eta\neq0$ a small real number). The corresponding perturbed solution is: $$\varphi(t;t_0,\tilde{y}_0)=(y_0+\eta)\,e^{a^2(t-t_0)}.$$ The difference between the perturbed and the original solution is: $$\Delta(t)=\varphi(t;t_0,\tilde{y}_0)-\varphi(t;t_0,y_0) =\eta\,e^{a^2(t-t_0)}.$$

Step 4 · Growth of the difference

Since $a\neq0$ we have $a^2>0$, so the exponent $a^2(t-t_0)\geq0$ for all $t\geq t_0$. Therefore: $$|\Delta(t)|=|\eta|\,e^{a^2(t-t_0)}\geq|\eta| \quad\text{for all }t\geq t_0,$$ and in fact $$|\Delta(t)|=|\eta|\,e^{a^2(t-t_0)}\to+\infty\quad\text{as }t\to+\infty.$$

Step 5 · Failure of the stability condition

Fix any $y_0$ and any $\varepsilon>0$. We must show that no $\delta>0$ can satisfy the Lyapunov stability requirement. Choose any $\eta$ with $0<|\eta|<\delta$. The condition requires $|\Delta(t)|<\varepsilon$ for all $t\geq t_0$. However: $$|\Delta(t)|=|\eta|\,e^{a^2(t-t_0)}.$$ Since $a^2>0$, there exists $T>t_0$ such that $$e^{a^2(T-t_0)}>\frac{\varepsilon}{|\eta|},$$ namely $$T=t_0+\frac{1}{a^2}\ln\frac{\varepsilon}{|\eta|}+1,$$ at which point $|\Delta(T)|=|\eta|\,e^{a^2(T-t_0)}>\varepsilon$. Hence, no matter how small $|\eta|>0$ is chosen (i.e.\ no matter how small $\delta$ is), the difference $|\Delta(t)|$ will eventually exceed $\varepsilon$. The Lyapunov stability condition fails.

Step 6 · Conclusion: the solution is unstable

$$\boxed{ \text{The solution }y(t)=y_0\,e^{a^2(t-t_0)} \text{ is \textbf{unstable} for all }a\neq0. }$$

Step 7 · Geometric interpretation

Since $a^2>0$, every solution $y(t)=y_0\,e^{a^2(t-t_0)}$ grows exponentially as $t\to+\infty$. Any two solutions with different initial values $y_0$ and $\tilde{y}_0=y_0+\eta$ diverge from each other at an exponential rate: $$|\Delta(t)|=|\eta|\,e^{a^2(t-t_0)}\to+\infty.$$ Trajectories starting arbitrarily close together are eventually driven arbitrarily far apart. This is the hallmark of instability.

Step 8 · Comparison with Problem 86

The sign of the coefficient in the exponent is the decisive factor: a negative coefficient produces decay and stability, while a positive coefficient produces unbounded growth and instability.

Summary Table

Additional Notes

• This result follows directly from the ε–δ definition of stability applied to the exact perturbation formula derived above; no linearisation or eigenvalue argument is needed for a scalar equation of this type.