The solutions are the ones you listed.
The solutions all have shape $y=(x-c)^2$ or $y=0$. Thus if $b<0$, then none of the solutions curves pass through $(a,b)$. So for all pairs $(a,b)$ such that $b<0$, there cannot be a solution satisfying $y(a)=b$. We do not know (yet) whether these are all the pairs $(a,b)$ for which there is no solution, but soon we will.
For any $a$, if $b=0$ there are exactly two solutions satisfying $y(a)=b$, the singular solution and the solution $y=(x-a)^2$.
Finally, we look at pairs $(a,b)$ with $b$ positive. We look for values of $c$ such that $y(a)=b$.
The solution $y=(x-c)^2$ passes through $(a,b)$ if and only if $(a-c)^2=b$. This equation has exactly two solutions, $c=a\pm\sqrt{b}$.
Conclusion: (a) The pairs $(a,b)$ for which there is no solution satisfying $y(a)=b$ are all $(a,b)$ with $b<0$. (b) There are no pairs $(a,b)$ for which there are infinitely many solutions with initial condition $y(a)=b$. (c) For all remaining pairs $(a,b)$, that is, all pairs with $b \ge 0$, there are finitely many solutions, indeed exactly two solutions that satisfy $y(a)=b$.
The geometry: The conclusion can also be reached geometrically, by visualizing the family of parabolas. All of your parabolas are obtained by sliding the standard parabola $y=x^2$ along the $x$-axis. For any $(a,b)$ with $b \gt 0$, there are exactly two such parabolas that pass through (a,b): one whose "left" half goes through $(a,b)$, and one whose "right" half goes through $(a,b)$.
Note: One could interpret the word "finite" to include the possibility of $0$ solutions: $0$ is certainly finite! That is obviously not the intended interpretation here. But if we interpret "finite" as including $0$, the answer to (c) is all pairs $(a,b)$.
Your differential system looks like this: $$\dot x=Ax\qquad x(0)=x_0$$ where $A$ is a $2\times 2$ matrix. Suppose that $x_0$ is an eigenvector of $A$. Then, when $t=0$, $\dot x=\lambda x_0$, so the derivative of the solution curve points in the direction of $x_0$. If $\lambda<0$, the curve is moving towards the origin. If $\lambda>0$, the curve is moving away from the origin.
These eigendirections give us a "skeleton" around which to build the geometry of the set of solutions to the differential system. If we have two different eigenvectors $v_1$ and $v_2$ with eigenvalues $\lambda_1$ and $\lambda_2$, (which your question implicitly assumes), then we can write any $x(t)$ as $$x(t)=a(t)v_1+b(t) v_2$$ for some scalar functions $a$ and $b$. Then $$\dot x=\dot a(t)v_1+\dot b(t) v_2=Ax=a(t)Av_1+b(t)Av_2=\lambda_1 a(t)v_1+\lambda_2 b(t) v_2$$ which, by the uniqueness of the decompositions in the base $(v_1,v_2)$, implies that $$\dot a(t)=\lambda_1a(t)\qquad \dot b(t)=\lambda_2b(t)$$ So the behavior of the solutions of the system is really dependent on the behavior along each eigenvector.
- If $\lambda_1<0<\lambda_2$, then one component, or one part of the solution is getting pulled toward the origin, while the other part is repelled. This leads to the saddle.
- If $\lambda_1< \lambda_2<0$, then both parts of the solution are pulled toward the origin, and we get a sink.
- If $0<\lambda_1< \lambda_2$, then both parts are pushed away from the origin, and so we get a source.
As for why we can tell all of this without knowing the specifics of the matrix, the answer is that if a matrix has two distinct eigenvalues, we know everything about the behavior of that matrix just from its eigenvalues and eigenvectors.
Best Answer
No. You consider a differential equation where the derivative is not given explicitly, i.e., your equation is $$ F(x,y,p)=0,\quad p=\frac{dy}{dx}. $$ The condition $F(x,y,p)=0$ gives you a surface $\pi$ in the space $(x,y,p)$. It is not difficult to figure out that your singular solution (which is actually called a discriminant curve) is given by the projection of the set of points of $\pi$, where the tangent plane is vertical.
A very readable account of the underlying theory can be found in Arnold's book, Section 1.3.