check for the following if they have a single solution on (-1,1)

A)

$\displaystyle y'=\sqrt{|y|}$

y(0)=0

B)

$\displaystyle y'=|y|^{3}$

y(0)=0

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- Nov 13th 2010, 06:25 AMtransgalacticsingle solution
check for the following if they have a single solution on (-1,1)

A)

$\displaystyle y'=\sqrt{|y|}$

y(0)=0

B)

$\displaystyle y'=|y|^{3}$

y(0)=0 - Nov 13th 2010, 10:20 AMAckbeet
In order to find a singular solution, it must be referenced to a family of solutions that you've already found. That is, a singular solution is a solution that can't be expressed as a member of a family of solutions. What is the family of solutions you've already found?

- Nov 13th 2010, 01:07 PMtransgalactic
for A :

$\displaystyle dy/dx=\sqrt{|y|}$

$\displaystyle \int \frac{dy}{\sqrt{|y|}}=\int dx$

i dont know how to deal with the absolute value

ill just egnore it

$\displaystyle 2y^{\frac{1}{2}}=x$

$\displaystyle 4y=x^2 +c$

by single solutioni didnt mean singluar solution which for 1/x x=0 is a singular point of solution

i ment like in lenear algebra where we get a single solution from a row redused matrice - Nov 13th 2010, 05:57 PMAckbeet
My mistake. I see what you mean now.

I think you're off by a factor of 2 in your solution. You should always plug your solution back into the DE to make sure it solves the DE. It's easy and it prevents mistakes. I also think that y = 0 is a perfectly good solution for both (A) and (B). Hence, I think both of them have multiple solutions. Indeed, the conditions of the standard existence theorem are not satisfied, so you're not guaranteed a unique solution. - Nov 13th 2010, 09:22 PMtransgalactic
you are correct i fixed the original post

so y=0 is the single solution

? - Nov 15th 2010, 01:46 AMAckbeet
I would say that y = 0 is a singular solution. You cannot get the solution y = 0 from any choice of C in the family of solutions $\displaystyle 4y=x^2 +c.$

There is more than one solution to the IVP's for both (A) and (B).