Starting from the definition of stability, examine the…

Starting from the definition of stability, examine the stability of the solution of the differential equation $$\frac{dx}{dt}=t-x,\quad x(0)=1.$$

Explanation of Method

The stability of the solution is examined directly from the ε–δ (Lyapunov) definition, without relying on any general theorem. The explicit solution is derived first; the initial condition is then perturbed by a small quantity, and the resulting difference between the perturbed and original solutions is estimated for all admissible values of the independent variable, and again in the limit as it tends to +∞.

Step 1 · Explicit solution

The equation is a first-order linear non-homogeneous ODE: $$x'+x=t.$$ Integrating factor: $\mu(t)=e^t$. Multiply both sides: $$(e^t x)'=te^t.$$ Integrate the right-hand side by parts ($\int te^t\,dt=te^t-e^t+C$): $$e^t x=te^t-e^t+C \implies x(t)=t-1+Ce^{-t}.$$ Apply $x(0)=1$: $1=-1+C$, so $C=2$. The solution is: $$\varphi(t;\,0,1)=t-1+2e^{-t}.$$

Step 2 · General solution with perturbed initial condition

For a general initial condition $x(0)=x_0$, applying $C=x_0+1$: $$\varphi(t;\,0,x_0)=t-1+(x_0+1)e^{-t}.$$ Let $\tilde{x}_0=x_0+\eta$ be a perturbed initial condition. Then: $$\varphi(t;\,0,\tilde{x}_0)=t-1+(\tilde{x}_0+1)e^{-t} =t-1+(x_0+\eta+1)e^{-t}.$$

Step 3 · Difference between perturbed and original solution

$$\Delta(t)=\varphi(t;\,0,\tilde{x}_0)-\varphi(t;\,0,x_0) =(x_0+\eta+1)e^{-t}-(x_0+1)e^{-t}=\eta\,e^{-t}.$$

Step 4 · Estimation of the difference

For all $t\geq0$: $$|\Delta(t)|=|\eta|\,e^{-t}\leq|\eta|,$$ since $e^{-t}\leq1$ for $t\geq0$. Moreover: $$\lim_{t\to+\infty}|\Delta(t)|=\lim_{t\to+\infty}|\eta|\,e^{-t}=0.$$

Step 5 · Verification of Lyapunov stability

Given any $\varepsilon>0$, choose $\delta=\varepsilon$. Then: $$|\tilde{x}_0-x_0|=|\eta|<\delta=\varepsilon \implies |\Delta(t)|=|\eta|\,e^{-t}\leq|\eta|<\varepsilon \quad\text{for all }t\geq0.$$ The Lyapunov stability condition is satisfied. Hence the solution is stable.

Step 6 · Verification of asymptotic stability

Since $e^{-t}\to0$ as $t\to+\infty$: $$\lim_{t\to+\infty}|\Delta(t)|=|\eta|\lim_{t\to+\infty}e^{-t}=0,$$ regardless of the size of $\eta$. Every perturbed solution converges to the original solution as $t\to+\infty$. Hence the solution is asymptotically stable.

Step 7 · Behavior of the original solution

The original solution $$\varphi(t)=t-1+2e^{-t}$$ consists of:

a particular solution $t-1$, which grows linearly — this is the dominant behavior for large $t$;
a transient term $2e^{-t}$, which decays exponentially to zero.

As $t\to+\infty$, $\varphi(t)\sim t-1$. All solutions with different initial values $x_0$ share the same asymptote $t-1$, since $(x_0+1)e^{-t}\to0$ for every $x_0$.

Step 8 · Geometric interpretation

All solutions $$\varphi(t;0,x_0)=t-1+(x_0+1)e^{-t}$$ are asymptotic to the line $x=t-1$ as $t\to+\infty$. The distance between any two solutions decays exponentially: $$|\Delta(t)|=|\eta|\,e^{-t}\to0.$$ Trajectories starting at different points converge toward one another, all funneling onto the same linear asymptote. This exponential contraction is the geometric manifestation of asymptotic stability.

Step 9 · Conclusion

$$\boxed{ \text{The solution }\varphi(t)=t-1+2e^{-t} \text{ is \textbf{asymptotically stable}.} }$$ Key observations:

The stability is uniform: $\delta=\varepsilon$ does not depend on $t_0=0$.
The decay rate of perturbations is $e^{-t}$, governed by the eigenvalue $\lambda=-1$ of the homogeneous part $x'+x=0$.
Since $\lambda=-1<0$, the homogeneous solution decays, and the particular solution $t-1$ serves as the common attractor for all trajectories.

Summary Table

Quantity	Value
Explicit solution	$x'+x=t.$
Perturbation difference	$\Delta(t)=\varphi(t;\,0,\tilde{x}_0)-\varphi(t;\,0,x_0) =(x_0+\eta+1)e^{-t}-(x_0+1)e^{-t}=\eta\,e^{-t}.$
Classification	asymptotically stable

Additional Notes

The stability is uniform: $\delta=\varepsilon$ does not depend on $t_0=0$.
The decay rate of perturbations is $e^{-t}$, governed by the eigenvalue $\lambda=-1$ of the homogeneous part $x'+x=0$.
Since $\lambda=-1<0$, the homogeneous solution decays, and the particular solution $t-1$ serves as the common attractor for all trajectories.

Result

The final classification is:

$$\boxed{ \text{The solution }\varphi(t)=t-1+2e^{-t} \text{ is \textbf{asymptotically stable}.} }$$