Question

Find the curves for which \(\frac{d y}{d x}=\frac{y^{2}+3 x^{2} y}{x^{2}+3 x y^{2}}\), and determine their orthogonal trajectories.

Short answer

In summary, to find the orthogonal trajectories of the given differential equation, follow these steps: 1. Solve the given differential equation: \(\frac{d y}{d x}=\frac{y^{2}+3 x^{2} y}{x^{2}+3 x y^{2}}\). 2. Rewrite the general solution in terms of the original variables: \(\frac{1}{2}[\frac{1}{\sqrt{3}}\arctan(\frac{y}{x\sqrt{3}})-\frac{1}{1+3(y/x)^2}] = \ln|x|+ C\). 3. Reverse the roles of \(x\) and \(-y\) in the given differential equation, and find the family of curves that satisfy this new differential equation: \(\frac{d x}{d y} = \frac{x^{2}+3 y^{2} x}{y^{2}+3 y x^{2}}\). By solving the new differential equation, you will find the orthogonal trajectories of the given differential equation.

Step-by-step solution

Solve the given differential equation

Rewrite the given differential equation as \(\frac{d y}{d x}=\frac{y^{2}+3 x^{2} y}{x^{2}+3 x y^{2}}\). Notice that this equation has the form of a homogeneous differential equation. We may transform this equation into a separable differential equation through substitution. Let \(y=vx\), then \(\frac{d y}{d x} = v + x \frac{d v}{d x}\). Substituting into the given equation, we have: $$v + x \frac{d v}{d x} = \frac{(vx)^{2}+3 x^{2} (vx)}{(x^2+3 x (vx)^2)}$$ Simplify the equation by dividing both numerator and denominator by \(x^2\): $$v + x \frac{d v}{d x} = \frac{v^2+3 v}{1+3 v^2}$$ Now subtract 'v' from both sides: $$x \frac{d v}{d x} = \frac{v^2+3 v}{1+3 v^2} - v$$

Get the family of curves from the general solution

Dividing both sides by \((1+3v^2)^2\), we have a separable equation: $$x \frac{d v}{d x} = \frac{v(1-v^2) }{(1+3 v^2)^2}$$ Separate the variables: $$\frac{1-v^2}{v(1+3v^2)^2}dv = \frac{1}{x}dx$$ Integrate both sides: $$\int \frac{1-v^2}{v(1+3v^2)^2} dv = \int \frac{1}{x} dx$$ The left integral can be solved using partial fractions or trigonometric substitution. After integrating, we get: $$\frac{1}{2}[\frac{1}{\sqrt{3}}\arctan(\frac{v}{\sqrt{3}})-\frac{1}{1+3v^2}] = \ln|x|+ C$$ Substitute back \(y = vx\): $$\frac{1}{2}[\frac{1}{\sqrt{3}}\arctan(\frac{y}{x\sqrt{3}})-\frac{1}{1+3(y/x)^2}] = \ln|x|+ C$$ This is the family of curves satisfying the given differential equation.

Find the orthogonal trajectories

To find the orthogonal trajectories, reverse the roles of \(x\) and \(-y\) and find a new differential equation: $$\frac{d x}{d y} = \frac{x^{2}+3 y^{2} x}{y^{2}+3 y_x^2}$$ Now, follow the same the steps as above by simplifying the equation and finding the family of curves that satisfy the new differential equation. The result of this process will be the orthogonal trajectories of the given differential equation.

Key concepts

Homogeneous Differential Equations

Understanding homogeneous differential equations is crucial in solving certain types of problems involving rates of change that can be modeled mathematically. These are specific types of differential equations that exhibit proportionality within the terms. Homogeneity refers to the property that all terms of the equation are of the same degree. In the context of our exercise, you might have wondered how identifying the equation as homogeneous helps in solving it. The trick lies in the substitution technique.

When we substitute a new variable, such as in our step by step solution with the relation \(y = vx\), we are leveraging the homogeneous nature of the equation to simplify it into a form that is easier to handle. Post-substitution, the equation no longer looks homogeneous—because we have fundamentally changed its structure into a separable form—but initially recognizing it as homogeneous guides us towards the appropriate solving method. The concept of homogeneous differential equations often intertwines with separable equations, as in our exercise, where the solving of one leads to the interpretation of the other.

Separable Differential Equations

Separable differential equations are another foundational concept in differential calculus. They allow the transformation of complex relationships into simpler, more solvable forms. As the term suggests, these equations can be separated into distinct parts that are only functions of \(x\) or \(y\). In our exercise, the transformation of the equation into a separable form was pivotal. It permitted us to express the two variables on separate sides of the equation, as seen in the solution after the substitution.

This separation pays off when we reach the integration stage. The ability to integrate both sides of the equation with respect to their respective variables (\(dv\) on one side and \(dx\) on the other) is an exclusive benefit of separable equations. This procedure effectively decouples the intertwined relationship between \(x\) and \(y\), or \(v\) in the substituted form, allowing for the equation to be solved through integral calculus which brings us to our next crucial concept.

Integral Calculus

The practice of integral calculus is fundamental to solving many types of differential equations, including separable ones. Once the differential equation is separated, as explained in the previous section, integration is used to find an anti-derivative for both sides, often resulting in the general solution to the differential equation.

In our example, we integrate to uncover the implicit relationship between \(x\) and \(v\), and hence \(x\) and \(y\). The integrals on both sides of the equation may not always look straightforward. Techniques such as partial fractions, trigonometric substitution, or integration by parts—depending on the form of the equation—are employed to solve complex integrals. The results can then be expressed in terms of inverse functions or logarithmic values as seen in the solution. Integral calculus not only provides the bridge from a differential equation to its solution but also equips us with the tools to mine deeper insights into the behavior of variables and their rates of change in numerous applications.