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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 $$~(y')^2= 4y,\quad y(a)= b~$$ 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.

2 Answers 2

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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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    @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 $(-1,7)$.2011-08-16