Question

If (dy/dx)= (1/√(y)), with y(0) = 9, find y and state its domain.

I found y= (3x/2 + 27)^(2/3), but I couldn't find the domain.

Answer 1

huh? do you need to integrate first?

Answer 2

is this really
\[\frac{dy}{dx}=\frac{1}{\sqrt{y}}\]?

Answer 3

yes

Answer 4

aaah ok, diffeq. got it.

Answer 5

off hand i can't see any restriction on the domain, but i will let someone else answer.

Answer 6

Your solution is correct.

Now, the function itself does in a (up to date) analysis of domain have domain over the entire reals. However, if we look at the original ODE, we must have
y > 0. In which case x > -18.

Answer 7

that what i got but my teacher disagreed

Answer 8

I don't know what to tell you. If y =< 0, the function doesn't satisfy the ODE because the ODE is non sensical.

Answer 9

i agree

Answer 10

@jamesj, suppose i have
\[f(x)=\sqrt[3]{x^2}\] with domain all real numbers. then
\[f'(x)=\frac{3}{2\sqrt[3]{x}}\] a function undefined at x = 0. but doesn't mean my original function is undefined there, it just means it is not differentiable there. i guess i am confused on this point.

Answer 11

now i'm really lost

Answer 12

My point is the the original ode we have this term:
\[ \frac{1}{\sqrt{y}} \]

That expression is only defined if y > 0.

Now there are two schools of thought about solutions of ODEs. One says the solutions must satisfy the ODE. The other says take those solutions and maximally extend them.

If you apply the first approach then we must have y > 0 and hence x > -18.

If you adopt the other approach, then the domain of x is the entire reals.

Answer 13

ah got it.
thanks