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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, y(a)= b has (a) no solution , (b) 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.

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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 $(a,b)$.

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Thank you so much for your reply. It would be of great help if you can please help me visualize the parts (b) and (c)geometrically. Because it seems to me graphically that in part (b), there are infinitely many solutions if b is greater than or equal to zero. – Tav Aug 16 '11 at 18:44
However, in part (c) if we move in delta neighborhood of point a, graphically it seems to be that we will have two distinct solutions. – Tav Aug 16 '11 at 18:56
@Tavleen: It could be partly a matter of interpreting the question, so do check exact wording. Right now it asks for all pairs $(a,b)$ for which there are infinitely many solutions $y(x)$ such that $y(a)=b$. So $(a,b)$ is temporarily fixed, like $(-1,7)$. Only two of your parabolas go through $(-1,7)$. Think of sliding the vertex of $y=x^2$ along the $x$-axis. We get your parabolas. Such a slid $y=x^2$ can go through $(-1,7)$ in only two ways: (i) the left half goes through $(-1,7)$; (ii) the right half goes through $(-1,7)$. – André Nicolas Aug 16 '11 at 19:02
thanks for your help . – Tav Aug 18 '11 at 18:05

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