Problem 2 - Technical Mathematics with Calculus

Question

Find the general solution of each differential equation. Try some by calculator. $$x d y=(4-y) d x$$

Step-by-Step Solution

Answer

The general solution of the differential equation \(x dy = (4-y) dx\) is \(y = 4 - A x\), where A is an arbitrary constant.

Step 1: Rearrange the differential equation

Isolate the terms involving y on one side and the terms involving x on the other side of the equation to facilitate separation of variables. We get \[\frac{dy}{dx} = \frac{4-y}{x}\].

Step 2: Separate variables

Rearrange the equation so all terms containing y are on one side and all terms containing x are on the other side. This gives us \[y - 4 = -x \frac{dy}{dx}\] or equivalently \[\frac{dy}{4-y} = -\frac{dx}{x}\].

Step 3: Integrate both sides

Integrate both sides of the equation with respect to their respective variables to find the solution. So we have \[\int \frac{1}{4-y}dy = -\int \frac{1}{x}dx\].

Step 4: Solve the integrals

The integral of 1 over (4 - y) with respect to y is \[\int \frac{1}{4-y}dy = -\ln|4-y| + C_1\] and the integral of -1 over x with respect to x is \[\int -\frac{1}{x}dx = -\ln|x| + C_2\].

Step 5: Combine constants and solve for y

Since both sides of the equation now contain a natural logarithm, we can combine the constant terms (\(C_1\) and \(C_2\)) into a single constant, \(C\). Then we represent the general solution in terms of y: \[ -\ln|4-y| = -\ln|x| + C \Rightarrow \ln|4-y| = \ln|x| - C \Rightarrow |4-y| = e^{-C} |x| \Rightarrow 4-y = \pm e^{-C} x\].

Step 6: Solve for the particular solution

Expose the arbitrary constant as a new constant \(A = \pm e^{-C}\), which can be any real number since C was an arbitrary constant. Isolate y to get the general solution: \[y = 4 - A x\].

Key Concepts

Separation of Variables

Separation of variables is a powerful method used when dealing with differential equations. It involves rearranging a differential equation so that all terms containing one variable are on one side of the equation, and all terms containing the other variable are on the other side. This step is crucial for tackling problems that can be solved by integrating both sides of the equation separately.

For example, consider the equation \[x dy = (4 - y) dx\]. The goal is to isolate the variables, which leads to \[\frac{dy}{dx} = \frac{4 - y}{x}\]. By further manipulation, we get \[\frac{dy}{4 - y} = -\frac{dx}{x}\], effectively separating the variables. The beauty of this method lies in its ability to reduce a potentially complex differential equation to simpler integrals that can be tackled individually.

Integrating Differential Equations

Once variables are separated, the next pivotal step in solving differential equations involves integrating both sides of the equation. Integration helps us to reverse the process of differentiation, essentially 'undoing' the derivative to find a function that describes the relationship between the variables involved in the equation.

In our example, the separated equation led to \[\int \frac{1}{4 - y}dy = -\int \frac{1}{x}dx\]. Integration of these terms yields log functions, specifically \[ -\ln|4 - y| + C_1\] for the left-hand integral and \[ -\ln|x| + C_2\] for the right-hand integral. Here, \(C_1\) and \(C_2\) represent the constants of integration, which are crucial to forming the general solution.

General Solution of Differential Equations

The general solution to a differential equation encapsulates all possible solutions that satisfy the equation. It includes an arbitrary constant—or constants—reflecting the indefinite nature of the integration process. After performing the integration, combining constants of integration results in a more consolidated expression that constitutes the general solution of the differential equation.

Upon solving the integrals in our example, we combined the constants into a single constant \(C\), leading to the equation \[|4 - y| = e^{-C} |x|\]. Isolating \(y\) completes the process, providing a general solution of the form \[y = 4 - A x\], where \(A = \pm e^{-C}\) encapsulates all possible values that the constant may take. This flexibility is what allows the general solution to capture the essence and various particular solutions of the differential equation based on different initial conditions or constraints.