Question

The general solution of the equation \((d y / d x)=\left(x^{2} / y^{2}\right)\) is: (A) \(x^{3}+y^{3}=c\) (B) \(x^{3}-y^{3}=c\) (C) \(x^{2}+y^{2}=c\) (D) \(x^{2}-y^{2}=c\)

Short answer

The general solution of the given ODE is \(x^3 - y^3 = C\), which matches option (B).

Step-by-step solution

Separate the variables

To find the general solution of the given ODE, we first need to separate the variables. This means rewriting the original equation such that we have all the terms involving \(y\) on one side and all the terms involving \(x\) on the other side. We have: \[\frac{dy}{dx} = \frac{x^2}{y^2}\] Now, multiply both sides by \(y^2\) and divide both sides by \(x^2\), giving: \(y^2\frac{dy}{dx} = x^2\)

Integrate both sides

Now, we need to integrate both sides of the equation with respect to \(x\). \[\int y^2 \frac{dy}{dx} dx = \int x^2 dx\] Since we're integrating with respect to \(x\), we can treat \(y^2\) as a constant and perform a substitution: let \(u = y(x)\), so \(\frac{du}{dx} = \frac{dy}{dx}\). Then, our equation becomes: \[\int y^2 \frac{du}{dx} dx = \int x^2 dx\] Now we can integrate both sides: \[\int y^2 du = \int x^2 dx\] \[y^3 = \frac{1}{3}x^3 + C\]

Solve for the general solution

To solve for the general solution, we will multiply both sides by 3 and move all terms to one side: \[3y^3 - x^3 = C'\] Where \(C'\) is another constant (\(C' = 3C\)). Now we can rewrite the equation in the form given in the options: \(x^3 - y^3 = -C'\) Thus, the general solution of the given ODE is: \[x^3 - y^3 = C\] Comparing with the given options, we find that the general solution matches option (B): (B) \(x^3 - y^3 = C\)

Key concepts

Separation of Variables

When tackling first-order differential equations, one powerful method we can employ is the separation of variables. This technique involves rearranging the equation so that each variable and its differential are on opposite sides. Imagine having a mathematical sentence where you want to gather all the related parts, such as nouns or verbs, together. Similarly, for differential equations, you would group all the parts containing one variable, say y, on one side of the equation and those with another variable, say x, on the other side.

Here's how it's done: take the original equation \(\frac{dy}{dx} = \frac{x^2}{y^2}\) and multiply each side by terms that will isolate \(dy\) and \(dx\) on separate sides. In our example, the equation becomes \(y^2 dy = x^2 dx\) upon rearrangement. This sets us up nicely to integrate each side in the following step, bringing us closer to solving the equation.

Integrating Factor

The concept of an integrating factor is a technique mainly used for linear first-order differential equations that aren't easily separated. The main idea is to multiply the entire equation by a carefully chosen function, which will enable us to rewrite the equation as an exact derivative of a product of functions. Think of an integrating factor as a mathematical 'key' that unlocks the equation, transforming it into a form that's easy to integrate.

For our present equation, however, we didn't need to use an integrating factor, since separation of variables was sufficient to solve it. Nevertheless, integrating factors are incredibly useful for more complex equations where separation of variables simply isn't possible. It often takes the form \(e^{\int P(x)dx}\), where \(P(x)\) is a function derived from the standard form of the equation \(\frac{dy}{dx} + P(x)y = Q(x)\).

First-order Differential Equations

Understanding first-order differential equations is crucial since they describe a plethora of phenomena in the natural world, from growth and decay processes to the behavior of electrical circuits. A first-order differential equation is one where the highest derivative involved is the first derivative. These equations can often be visualized as representing rates of change, such as velocity being the rate of change of position with respect to time.

In solving a first-order differential equation like \(\frac{dy}{dx} = \frac{x^2}{y^2}\), our primary goal is to find a function \(y(x)\) that satisfies the equation. Solutions can take various forms, including a general solution that encompasses a family of solutions differentiated by constants or a particular solution that satisfies an additional condition, such as an initial value.

Integration of Functions

The process of integration of functions is a critical tool for solving differential equations. It's the reverse operation of differentiation, analogous to assembling a disassembled mechanism. When integrating, we're essentially summing infinitesimally small pieces to find the total quantity, whether it's the area under a curve or the antiderivative of a function.

In the solution to our differential equation exercise, after separating the variables, we carried out the integration on both sides - \(\int y^2 dy\) and \(\int x^2 dx\). These integrations yielded \(y^3/3\) and \(x^3/3\) plus the constant of integration, which allowed us to solve for \(y\) as a function of \(x\), thus finding the general solution to the differential equation.