A little help with calculus terms.

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The discussion centers on formulating and solving a differential equation based on the statement that the rate of change of y with respect to x is inversely proportional to the square root of y. The correct differential equation is identified as dy/dx = k * y^(-1/2). Participants clarify that the initial confusion regarding the presence of x in the equation was an error. The equation is recognized as separable, leading to the integration of y^(1/2) dy = k dx to find y as a function of x. The conversation emphasizes the importance of correctly interpreting the terms in the differential equation.
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The rate of change of y with respect to x is inversely proportional to the square root of y.
a)Write a differential equation for the given statement
b)Solve the differential equation in part a.

I don't know, but what I've done so far is:
({dy/dx}) k=y^{1/2}
 
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I'm not sure, but I think this is it:

a) (dy/dx)x=1/sqrt(y)

b)
(dy/dx)1/x²=y
-2x/x^4=y
 
Incorrect.It's INVERSE PROPORTIONALITY.We usually let the constant in the other side of the equality.

dy(x)/dx~y^{-\frac{1}{2}}=>\frac{dy(x)}{dx}=ky^{-\frac{1}{2}}

Daniel.
 
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ToxicBug said:
I'm not sure, but I think this is it:

a) (dy/dx)x=1/sqrt(y)

b)
(dy/dx)1/x²=y
-2x/x^4=y

Sorry,there's no "x" explicitely.Just "y" to a power & its first derivative of "y".


Daniel.
 
why is the x beside the (dy/dx)?
 
Nevermind, I think I misunderstood the whole point of the question :/
 
ok, but why is the "x" in the left side of the equation, why is there an x at all?
 
It isn't.It shouldn't be.It was an error from the poster.

Daniel.
 
so the solution should just be \frac{dy}{dx}=ky^{-\frac{1}{2}} ?
 
  • #11
that would mean that y=?
 
  • #12
You can find easier "x" as a function of "y".

Daniel.
 
  • #13
\frac{dy}{dx}= ky^{\frac{-1}{2}}
is a "separable equation". Write it as
y^{\frac{1}{2}}dy= kdx
and integrate.
 
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