Solving the Discontinuity of y=sqrt(25-x^2)

[vipertongn]

Homework Statement

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

 

[malicx]

Homework Statement

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

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.

 

[vipertongn]

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

 

[Mark44]

Homework Statement

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

The function above isn't continuous at x = 0.

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...

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.

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

Please give us the exact wording of the problem, and we can go from there.

 

[vipertongn]

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)

 

[Mark44]

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).