Find all solutions to the following initial value problem, or show that no solution exists:
$\begin{align} y^{\prime}=\left\{\begin{array}{ll} 3 x^2-1 & \text { if } x<-1 \\ 0 & \text { if } x \geq-1 \end{array} \quad \text { with } y(-2)=1\right. \end{align}$
How do I even begin to solve this problem? Our textbook also gives this theorem, which I tried applying (not sure if it can actually be applied here) to show that a solution does not exist:
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Theorem 1.4.4: Existence
If $F(x,y)$ is continuous for all $(x,y)$ near $(x_0, y_0)$, then the initial value problem $$y’=F(x,y), \quad (x_0)=y_0$$ Has a solution $y=y(x)$ for $x$ near $x_0$
but since this is only an introductory calculus course, we do not really prove the theorem, nor do we work with a rigid notion of "nearness". I tried graphing $y'$ and applying the theorem to show that $y'$ is discontinuous at $x=-1$, but did not really know "how near" would be "near enough" for "$x$ near $x_0$".
Direct integration of the ODE yields $$ y(x)=\begin{cases} x^3-x+c_1&\text{if $x<-1$,} \\ c_2&\text{if $x\geq-1$.} \end{cases} \tag{1} $$ The initial condition $y(-2)=1$ implies $(-2)^3-(-2)+c_1=1$, or $c_1=7$. To determine $c_2$, we impose continuity of $y(x)$ at $x=-1$: $$ \lim_{x\to-1^{-}}y(x)=\lim_{x\to-1^{+}}y(x)\implies c_1=c_2 \implies c_2=7. \tag{2} $$ In conclusion, $$ y(x)=\begin{cases} x^3-x+7&\text{if $x<-1$,} \\ 7&\text{if $x\geq-1$.} \end{cases} \tag{3} $$