Question

Solve the given initial value problem and determine at least approximately where the solution is valid. $$ (2 x-y) d x+(2 y-x) d y=0, \quad y(1)=3 $$

Short answer

Question: Determine the solution of the initial value problem (2x - y)dx + (2y - x)dy = 0 with the initial condition y(1) = 3 and the interval where the solution is valid. Answer: The solution of the initial value problem is given by the function Ψ(x,y) = x^2 - yx + y^2 = 5. The solution is valid for all (x,y) ∈ ℝ².

Step-by-step solution

Checking if the given DE is an exact differential equation

To check the exactness of the given DE, we must find the partial derivatives M_x and N_y and compare them. Here, M(x,y) = 2x - y and N(x,y) = 2y - x. Calculate the partial derivatives: $$ M_x = \frac{\partial}{\partial x} (2x - y) = 2 \\ N_y = \frac{\partial}{\partial y} (2y - x) = 2 \\ $$ Since M_x = N_y, the given DE is an exact differential equation.

Solving the exact differential equation

We can find a function Ψ(x,y) such that Ψ_x = M and Ψ_y = N. Let's find Ψ(x,y): Integrate M with respect to x: $$ \int(2x - y)dx = x^2 - yx + h(y) \quad(1) $$ where h(y) is an arbitrary function of y. To find h(y), differentiate equation (1) with respect to y: $$ \frac{\partial}{\partial y}(x^2 - yx + h(y)) = -x + h'(y) \quad(2) $$ Comparing equation (2) with N(y), we get: $$ -x + h'(y) = 2y - x \\ h'(y) = 2y $$ Now integrate h'(y) to find h(y): $$ h(y) = \int(2y)dy = y^2 + C $$ where C is the constant of integration. Now, we can write the solution Ψ(x,y) as follows: $$ Ψ(x,y) = x^2 - yx + y^2 + C $$ The solution of the exact differential equation is given by Ψ(x,y) = k, where k is a constant.

Using the initial condition

We are given that when x = 1, y = 3. Using this initial condition, we can find the value of the constant k in the solution: $$ Ψ(1,3) = k \\ 1^2 - 3(1) + 3^2 = k \\ k = 5 $$ So the particular solution for the given DE and initial condition is Ψ(x,y) = 5, which can be written as: $$ x^2 - yx + y^2 = 5 $$

Determining the valid interval for the solution

To find the valid interval, we can look for the region in the xy-plane where the solutions exist. In our case, since there is no restriction on the domain of the solution function, the solution is valid for all (x,y) ∈ ℝ².

Key concepts

Initial Value Problem

An Initial Value Problem (IVP) is a specific type of differential equation accompanied by a condition termed as the initial condition. This condition specifies the value of the unknown function at a particular point, often starting time or position. The purpose of solving an IVP is to find a function that not only satisfies the differential equation but also fulfills the initial condition, thus providing a unique solution to the problem.

To illustrate, let's consider the exercise problem. We have a differential equation and the initial condition given as y(1) = 3. The goal is to find a function y = f(x) that solves the differential equation for each value of x and that passes through the point (1,3). In the step-by-step solution provided, this process involved integrating and finding the constant k thanks to the initial value, leading to a specific equation that describes the behavior of the system described by the original differential equation.

Partial Derivatives

In calculus, Partial Derivatives are used when dealing with multiple variables and we want to measure the rate at which one variable changes with respect to another, keeping all other variables constant. They are fundamental in solving exact differential equations as well as finding local extrema of multivariable functions.

Considering our exercise, the partial derivatives M_x and N_y of the functions M(x, y) and N(x, y), respectively, are computed. If these partial derivatives are equal, as they are for the given case, the differential equation is deemed exact. It implies that there exists a function Ψ(x, y) whose differential matches the left-hand side of our original equation. In more complex settings, identifying if an equation is exact by comparing partial derivatives can save one from unnecessary trials of non-applicable solution methods.

Integration of Functions

The term Integration of Functions refers to the process of finding a function when its rate of change, or its derivative, is known. Integration is a core concept in mathematics that allows us to determine quantities like areas, volumes, and, in the context of differential equations, potential functions from their gradients.

In the exercise problem, we use integration to find the function Ψ(x, y) from the given rate-of-change functions M(x, y) and N(x, y). Integrating M with respect to x and N with respect to y and then combining the results gives us the potential function, which is the solution to the exact differential equation. The integration step is crucial, and the selection of integration constants is often refined by the initial condition, yielding the particular solution that describes the behavior of the system.