Question

Find a first order differential equation for the given family of curves. $$ y\left(x^{2}+y^{2}\right)=c $$

Short answer

In this exercise, we were given a family of curves represented by the equation $$y\left(x^{2}+y^{2}\right)=c$$, where c is a constant. To find the first-order differential equation that represents this family of curves, we used implicit differentiation and followed these steps: 1. Differentiate both sides of the equation $$y\left(x^{2}+y^{2}\right)=c$$ with respect to x. 2. Apply the product rule and chain rule to differentiate the left side of the equation. 3. Rearrange the equation and solve for \({dy}/{dx}\). Following these steps, we found the first-order differential equation to be: $$\frac{dy}{dx} = \frac{-2xy}{x^{2}+3y^{2}}$$.

Step-by-step solution

Implicit Differentiation of the Given Equation

To find the first-order differential equation from the given equation, we need to differentiate both sides of the equation $$y\left(x^{2}+y^{2}\right)=c$$ with respect to x. Since x and y are related, we will use implicit differentiation. We will differentiate both sides with respect to x and solve for \({dy}/{dx}\) (the first-order derivative of y with respect to x). Differentiating both sides with respect to x:$$ \frac{d}{dx}\left[y\left(x^{2}+y^{2}\right)\right] = \frac{d}{dx} c $$

Differentiate the Left Side and Rearrange

By applying the product rule and chain rule, we can differentiate the left side of the equation: $$ \begin{aligned} \frac{d}{dx}\left[y\left(x^{2}+y^{2}\right)\right] &= \frac{dy}{dx}\left(x^{2}+y^{2}\right) + y\frac{d}{dx}\left(x^{2}+y^{2}\right)\\\\ &= \frac{dy}{dx}\left(x^{2}+y^{2}\right) + y(2x+2y\frac{dy}{dx}) \end{aligned} $$Since the derivative of a constant is zero, the right side of the equation becomes:$$ \frac{d}{dx} c = 0 $$Now, we have:$$ \frac{dy}{dx}\left(x^{2}+y^{2}\right) + y(2x+2y\frac{dy}{dx}) = 0 $$

Solve for \({dy}/{dx}\)

Rearrange the equation and factor out \({dy}/{dx}\) to solve for the first-order differential equation:$$ \begin{aligned} \frac{dy}{dx}\left(x^{2}+y^{2}\right) + y(2x+2y\frac{dy}{dx}) &= 0\\\\ \frac{dy}{dx}(x^{2}+y^{2}+2y^{2}) &= -2xy\\\\ \frac{dy}{dx} &= \frac{-2xy}{x^{2}+3y^{2}} \end{aligned} $$Now, we have found the first-order differential equation:$$ \frac{dy}{dx} = \frac{-2xy}{x^{2}+3y^{2}} $$

Key concepts

Implicit Differentiation

Implicit differentiation is a powerful technique used when dealing with equations where the variable 'y' is not isolated and both 'x' and 'y' are mingled. For instance, consider the equation of a family of curves provided in the exercise: \( y(x^2 + y^2) = c \). It is not easy to solve for 'y' and then differentiate, so we apply implicit differentiation to find the derivative of 'y' with respect to 'x' directly from the given equation.

Here's how it works: we will differentiate each term with respect to 'x', treating 'y' as a function of 'x'. This approach recognizes that even though 'y' depends on 'x', we can still work through the equation term-by-term. For example, when we differentiate 'y' times something, where that something involves 'x', we will apply the product rule, another key concept, to take the derivative correctly.

Product Rule

When facing a term in which two functions are multiplied together, like 'y' and \(x^2 + y^2\), we can't simply differentiate each individually; we have to apply the product rule. The product rule states that the derivative of a product of two functions is the derivative of the first function multiplied by the second function, plus the first function multiplied by the derivative of the second function.

Following the product rule, we take the derivative of 'y' as \(\frac{dy}{dx}\) since it's with respect to 'x', and keep \(x^2 + y^2\) intact. Then, we add to this the original 'y' multiplied by the derivative of \(x^2 + y^2\), which we find by using the chain rule. The product rule is essential when variables are intertwined within a product, ensuring we account for the variation in both when differentiating.

Chain Rule

The chain rule is a formula for computing the derivative of the composition of two or more functions. Put simply, if one variable depends on another, which in turn depends on a third variable, the chain rule helps us to differentiate this compound relationship. In our exercise, we use the chain rule to differentiate \(x^2 + y^2\).

When differentiating this term with respect to 'x', we treat \(y^2\) as a composition. Hence, for the \(x^2\) term, the derivative is straightforward: \(2x\). But for \(y^2\), we must acknowledge that 'y' is a function of 'x' and apply the chain rule: first finding the derivative of \(y^2\) as if 'y' were the variable, which is \(2y\), and then multiplying this by the derivative of 'y' with respect to 'x' or \(\frac{dy}{dx}\). This compound derivative is crucial to correctly differentiate expressions where one variable is expressed as a function of another, especially in the context of implicit differentiation.