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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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up vote 8 down vote accepted

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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Can you please expand on the following steps please. 3x^2 + 3y^2\left(\frac{dy}{dx}\right) &= \frac{d}{dx}3xy &&\text{(using the Chain Rule)}\\ when you use Chain rule, why do you leave the 3y^2\left(\frac{dy}{dx}\right) in this when you didn't do that for the first. –  user10695 Jun 16 '11 at 21:55
    
@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. –  Arturo Magidin Jun 17 '11 at 3:08
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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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Alas, I am too late and too redundant! –  The Chaz 2.0 Jun 15 '11 at 5:02
    
When I took the derivative of $$x^2 + y^2 = 9$$ , I got $2x+2y =0$ instead. May I ask how you got your answer please? –  user10695 Jun 15 '11 at 5:06
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He used the chain rule, like he said! You could write, after differentiating both sides with respect to x, $2xx' + 2yy' = 0$, but keep in mind that $x' = \frac{dx}{dx} = 1$, so that's why it doesn't appear in his answer. –  The Chaz 2.0 Jun 15 '11 at 5:08
    
@The Chaz: I was elsewhere at the time. You got it. Thanks. –  Ross Millikan Jun 15 '11 at 5:10
    
You were probably beating me to the punch elsewhere! :) –  The Chaz 2.0 Jun 15 '11 at 5:16
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