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I need to determine whether the function defined implicitly by: $x^2 + y^2 = 9$ is a solution of the differential equation: $\frac{dy}{dx}= \frac{x}{y}.$ Please explain each step in process. I would like to very much understand the solution.

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You don't actually have to solve the differential equation to verify that the given function is a solution. What you need to do is "plug in" the proposed solution into the differential equation and see if you get the right answer. If you do, then it's a solution. If you don't, then it's not a solution.

But this is not "really" a differential equation anyway. What you need to do is compute the derivative of $y$ with respect to $x$ using implicit differentiation.

To do that, just differentiate with respect to $x$ both sides of the equation, using the usual differentiation rules; but you must remember that $y$ is "really" a function of $x$, so that you should use the Chain Rule whenever you are differentiating an expression involving $y$. Since you don't know what y' is explicitly, you leave it indicated.

Once you are done, you "solve for $y'$".

Let me do a different example so you can see how it is done. Suppose you want to find $\frac{dy}{dx}$ for the function defined implicitly by $x^3+y^3 = 3xy.$ (The folium of Descartes) First, we take derivatives on both sides, remembering to use the Chain Rule; I'll do each term separately: \begin{align*} x^3 + y^3 &= 3xy\\ \frac{d}{dx}\left( x^3+y^3\right) &= \frac{d}{dx}3xy\\ \frac{d}{dx}x^3 + \frac{d}{dx}y^3 &= \frac{d}{dx}3xy &&\text{(since }(f+g)'=f'+g'\text{)}\\ 3x^2 + \frac{d}{dx}y^3 &= \frac{d}{dx}3xy &&\text{(since }(x^3)'=3x^2\text{)}\\ 3x^2 + 3y^2\left(\frac{dy}{dx}\right) &= \frac{d}{dx}3xy &&\text{(using the Chain Rule)}\\ 3x^2 + 3y^2y' &= \frac{d}{dx}3xy &&\text{(we don't know what }\frac{dy}{dx}\text{ is, so leave indicated)}\\ 3x^2 + 3y^2y' &= 3\frac{d}{dx}(xy) &&\text{(since }(3f)' = 3f'\text{)}\\ 3x^2+3y^2y' &= 3\left( x'y + xy'\right) &&\text{(product rule)}\\ 3x^2 +3y^2y' &= 3(y+xy') &&\text{(since }x'=1\text{)}\\ 3x^2 + 3y^2y' &=3y + 3xy' \end{align*} Now, solve for y': \begin{align*} 3x^2 + 3y^2y' &= 3y+3xy'\\ 3y^2y' - 3xy' &= 3y - 3x^2\\ 3(y^2-x)y' &= 3(y-x^2)\\ y' &= \frac{3(y-x^2)}{3(y^2-x)}\\ y' &= \frac{y-x^2}{y^2-x}. \end{align*} So for the Folium of Descartes, $\frac{dy}{dx} = \frac{y-x^2}{y^2-x}$.

That's what you need to do to see whether $\frac{dy}{dx}$ for the circle of radius $3$ and center at the origin is equal to $\frac{y}{x}$; but it will be much easier than the above, so if you can follow the derivation here, your problem should be a snap.

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    @user10695: $y$ is really a function of $x$; if we write it as $f(x)$ (instead of $y$), then it becomes clearer. We have $\frac{d}{dx}(f(x))^2$; how is this done? By using the Chain Rule: $\frac{d}{dx}(f(x))^2 = 2(f(x))*f'(x)$. Since $y=f(x)$, then $f'(x) = y' = \frac{dy}{dx}$. In fact I **did** do it "for the first", it's just that in that case you get $\frac{dx}{dx}$, which equals $1$, and of course multiplying by $1$ is the same as doing nothing.2011-06-17
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Do you know implicit differentiation? You can differentiate using the chain rule giving 2x+2yy'=0 and see if it satisfies y'=\frac{x}{y}

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    You were probably beating me to the punch elsewhere! :)2011-06-15