# Why is the piecewise defined function isn't a solution of the differential equation?

1. Aug 7, 2010

### vipertongn

1. The problem statement, all variables and given/known data
y = sqrt(25-x^2) -5<x<0
y = -sqrt(25-x^2) 0=<x<5

dy/dx= -x/y

Why is this not a solution on(-5,5) when...

y = sqrt(25-x^2)
y = -sqrt(25-x^2)
are solutions? kinda confused...

Could someone explain to me step by step how to solve this...I spent an hour trying to understand what they are asking for and solving the problem...haven't gotten anywhere

2. Aug 7, 2010

### malicx

Re: Why is the piecewise defined function isn't a solution of the differential equati

I'm confused as well, you didn't state a problem. Are you trying to show that

dy/dx = -x/y

is satisfied by the given equations? If so, it's very straightforward... differentiate (implicitly) the given equations and you can see it has the same form.

3. Aug 7, 2010

### vipertongn

Re: Why is the piecewise defined function isn't a solution of the differential equati

I want to know why it isn't a solution in that piecewise combination

4. Aug 7, 2010

### Staff: Mentor

Re: Why is the piecewise defined function isn't a solution of the differential equati

The function above isn't continuous at x = 0.
You're leaving out some information, it seems. The general solution of the DE dy/dx = -x/y is y = +/-sqrt(C - x^2). An initial condition, which you don't include, would enable you to solve for C, and determine which square root is the unique solution.
Please give us the exact wording of the problem, and we can go from there.

5. Aug 7, 2010

### vipertongn

Re: Why is the piecewise defined function isn't a solution of the differential equati

this is what it exactly says... that
y=phi1(x)=sqrt(25-x^2) and y=phi2(z)=-sqrt(25=x^2) are solutions of dy/dx=-x/y on the interval (-5,5). nd explain why the piece wise solution

y = sqrt(25-x^2) -5<x<0
y = -sqrt(25-x^2) 0=<x<5

isn't a solution of the DE on the interval (-5,5)

6. Aug 8, 2010

### Staff: Mentor

Re: Why is the piecewise defined function isn't a solution of the differential equati

Any solution of the diff. equation has to be differentiable at every point in the open interval (-5, 5). Your piecewise-defined function is neither continuous nor differentiable at x = 0. The two functions phi1(x) and phi2(x) are continuous and differentiable at every point of (-5, 5).