Using definition of Lyapunov's stability, check if following solution is stable: $$ \begin{cases} & x'(t)=-x + t^2\\ & x(1) = 1 \end{cases} $$
Now, I've solved the equation as I felt it was necessary:
$$ x' + x = t^2 \\ (xe^t)'=e^tt^2 \\ xe^t = e^t(t^2-2t+2) \\ x = t^2 -2t + 2 + ce^{-t} $$ Plugging the initial condition $x(1)=1$ we get $c = 0$. Solution is then $$ x(t) = t^2 -2t + 2 $$
I wish I could show anything more but I'm banging my head against the wall and can't find any meaningful examples what to do next.
My definition of Lyapunov's stability was given as:
Stationary point $x_0$ is called stable when $\forall_{\epsilon > 0} \exists_{\delta > 0} \|x-x_0\| < \delta \implies$
- $\varphi( ., x)$ is defined on $[0, \infty)$
- $\|\varphi(t,x) - x_0\| < \epsilon \forall t \geq 0$.
In view of the fact that you have been given the definition of stability of the equilibrium point, not stability of a solution in general, you are apparently expected to make a change of variables. Stability of the solution $$ x_s(t)=t^2-2t+2 $$ is equivalent to stability of equilibrium point $y_s=0$ of the system in variable $y(t)$, $y(t)=x(t)-x_s(t)=x(t)-t^2+2t-2$. The system in $y$ is $$ y'= x'-2t+2=-x+t^2-2t+2=-y. $$ The solution of the initial value problem $$ y'=-y,\quad y(0)=y_0 $$ is $y=y_0 e^{-t}$. Its norm decreases for any $t$, thus, you can take $\delta=\epsilon$ in the definition to prove the stability of $y(t)=0$.