Question

Solve the linear fractional equation $$ y^{\prime}=\frac{a x+b y}{c x+d y}, \quad a d \neq b c $$ Also obtain solutions by using polar coordinates.

Short answer

#Short Answer# The implicit general solution of the given linear fractional differential equation can be expressed as: $$ k = \int \frac{dy}{ax + by} - \int \frac{dx}{c + dy} $$ In polar coordinates, the solution is: $$ k = \int \frac{d(r\sin(\theta))}{a (r\cos(\theta)) + b (r\sin(\theta))} - \int \frac{d(r\cos(\theta))}{c + d(r\sin(\theta))} $$

Step-by-step solution

Rewrite the given equation

We are given the equation: $$ y^{\prime}=\frac{a x+b y}{c x+d y}, \quad a d \neq b c $$ Let's rewrite it in a more manageable way: $$ y^{\prime}(x) - \frac{a x + b y}{cx + dy} = 0 $$

Introduce a new dependent variable

Let's introduce a new dependent variable \(v(x) = cx + dy\). So, we can rewrite the given equation in terms of this new dependent variable: $$ y^{\prime}(x) - \frac{a x + b y}{v} = 0 $$

Differentiate the new dependent variable

Now, we will differentiate the new dependent variable with respect to x: $$ v'(x) = c + dy'(x) $$

Substitute the derivative into the given equation

Now let's substitute the expression for \(v'(x)\) into the given equation: $$ y^{\prime}(x) - \frac{a x + b y}{c + dy'(x)} = 0 $$

Solve the given equation

Now we will solve the given equation, which is now a separable ODE: $$ \bigg(c + dy'(x)\bigg)y^{\prime}(x) - (a x + b y) = 0 $$ We can separate the variables and integrate both sides: $$ \int \frac{dy}{ax + by} = \int \frac{dx}{c + dy} $$

Obtain the general solution

Integrating both sides, we get the implicit general solution: $$ k = \int \frac{dy}{ax + by} - \int \frac{dx}{c + dy} $$ where \(k\) is the integration constant.

Express the solution using polar coordinates

To express the solution using polar coordinates, we will use the following conversion formulas: $$ x = r\cos(\theta), \quad y = r\sin(\theta) $$ Substituting these into the general solution: $$ k = \int \frac{d(r\sin(\theta))}{a (r\cos(\theta)) + b (r\sin(\theta))} - \int \frac{d(r\cos(\theta))}{c + d(r\sin(\theta))} $$ This is the implicit general solution expressed in polar coordinates.

Key concepts

separable ODE

A separable ordinary differential equation (ODE) is one that can be expressed in the form \( \frac{dy}{dx} = g(x)h(y) \) where the variables can be separated onto opposite sides of the equation, making it possible to integrate each side with respect to its own variable. In other words, the equation can be manipulated to look like \( f(y)dy = g(x)dx \).

This is particularly useful because it simplifies the solving process, allowing us to find the function \( y(x) \) that satisfies the original equation. The solution to a separable ODE can often be expressed implicitly, where \( y \) is a function of \( x \) but is not explicitly solved for \( y \). In the given problem, after introducing a new dependent variable, the equation is transformed into a separable form, enabling us to find an implicit general solution through integration.

polar coordinates

Polar coordinates offer an alternative way to describe the position of a point, which is particularly handy in certain mathematical problems, such as those involving circular or rotational symmetry. Instead of defining a location in terms of \( x \) and \( y \) as in Cartesian coordinates, polar coordinates use the distance from the origin \( r \) and the angle \( \theta \) with respect to the positive x-axis.

The polar coordinate system is defined by the two equations \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \). In contexts like solving differential equations, converting from Cartesian to polar coordinates can sometimes simplify the problem, as it did in our exercise where the polar form allowed us to find a unique expression for the implicit general solution.

implicit general solution

The implicit general solution of a differential equation is a solution that is not explicitly solved for the dependent variable—often \( y \) as a function of the independent variable \( x \). Instead, it represents the relationship between \( y \) and \( x \) without isolating \( y \) on one side of the equation. In many cases, finding an explicit solution may be difficult or impossible, but an implicit solution can still provide valuable insights into the behavior of the system described by the differential equation.

In our specific problem, the general solution involves an integration constant and integrates both variables separately, ending up with an equation that depicts the relationship between \( y \) and \( x \) without giving a formula for \( y \).

differential equations

Differential equations are mathematical equations that relate some function with its derivatives. They describe various phenomena such as physics, engineering, biology, and economics. The order of a differential equation corresponds to the highest derivative present in the equation. Solving a differential equation involves finding the function or functions that satisfy the relationship specified by the equation.

In our textbook problem, we are dealing with a linear fractional differential equation which combines the derivatives of a function with a ratio of linear combinations of the function and its independent variable. These types of equations can often model more complex systems and require advanced techniques to solve, including variable substitution and transformation of coordinate systems like converting to polar coordinates as demonstrated in the step-by-step solution.