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.
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
|
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. |
|||||
|
|
Do you know implicit differentiation? You can differentiate using the chain rule giving $2x+2yy'=0$ and see if it satisfies $y'=\frac{x}{y}$ |
|||||||||||||
|
