Solve the differential equation $$~(y')^2= 4y~$$ to verify the general solution curves and singular solution curves. Determine the points $~(a,b)~$ in the plane for which the initial value problem $$~(y')^2= 4y,\quad y(a)= b~$$ has
$(a)\quad$ no solution ,
$(b)\quad$ infinitely many solutions
that are defined for all values of $~x (c)~$ on some neighborhood of the point $~x=a~$ , only finitely many solutions.
General solution that I am getting is $~y (x) = (x-c)^2~$ and singular solution is $~y(x)=0~$.
I wish a clarification in the second part.
The family of curves consists on parabolas with vertex varying on the $~x~$-axis.
I need someone to explain how to proceed to find suitable choice of $~(a,b)~$ in each part.
Best Answer
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 $(a,b)$.