A problem in first order ODE? I was reading the undergraduate level text book "Differential Equations and its Applications" by Martin Braun. 
Prove that $y(t)=-1$ is the only solution of the IVP
$y'=t(1+y), y(0)=-1$
I reasoned as follows:
As the $y'(0)=0$, the y never moves(I have Euler's scheme in mind) and therefore $y(t)=-1 \forall t$
Now I do not know how to explicitly compute this solution. Is there a way we could do that?
Clearly the separation of variable technique would not work as $\frac{\partial f}{\partial y}$ is not defined at the $t=0$
Though the author briefly explained it it, I am still a bit unclear as to what happens in such a situation.
When can an IVP have more than on solution and why are multiple solutions not acceptable in applications(the author mentions this without giving any explanation)
 I was thinking of applying  the following theorem to prove uniqueness:
Theorem:If $f$ and$\frac{\partial f}{\partial y}$ be continuous in a rectangle $R:t_0\leq t \leq t_0 +a,|y-y_0|\leq b$
and we compute 
$M=max_{(t,y) \in R}|f(t,y)|$ and we set $\alpha=min \big(a,\frac{b}{M}\big)$
Then the IVP $y'=f(t,y),\hspace{1 mm} y(t_0)=y_0$ has a unique solution $y(t)$ on the interval $t_0 \leq t \leq t_0 +\alpha$
Its just been a couple of weeks that I began studying ODE's so my question might not be clearly formulated but I tried to do my best
 A: The solution with $y(0)=-1$ is $y(t)=-1$ due to the Cauchy-Lipschitz theorem, uniqueness part. Assuming that $y(0)\neq -1$, in a neighbourhood of the origin we have $y(t)\neq 1$ and:
$$\frac{y'}{1+y}= t, \tag{1}$$
hence integrating from $0$ to $x$ ($x$ in such a neighbourhood) we get:
$$ \log(1+y(x)) = x^2/2 + C, $$
$$ y(x) = K e^{x^2/2} - 1, \tag{2}$$
with $K = y(0)+1$. The local solution so found is, indeed, a global solution for $(1)$.
A: We have $y^{\prime}-ty=t$, which is a 1st-order linear DE.  Since $P(t)=-t$ and $Q(t)=t$ are continuous in an open interval containing 0, the initial value problem has a unique solution.  
Since $y(t)=-1$ clearly satisfies the DE and the initial condition, it must be the unique solution.

We can also show this by solving the linear DE directly:
The integrating factor is given by $\mu(t)=e^{\int -t dt}=e^{-\frac{t^2}{2}}$, so multiplying both sides by  $\mu(t)$ gives
$\left(e^{-\frac{t^2}{2}}y\right)^{\prime}=te^{-\frac{t^2}{2}}\;\;$ and $\;\;e^{-\frac{t^2}{2}}y=\int te^{-\frac{t^2}{2}} dt=-e^{-\frac{t^2}{2}}+C$
Then $y=-1+Ce^{\frac{t^2}{2}}$, and $y(0)=-1\implies C=0\implies y=-1$.
