Question

Finding a Particular Solution In Exercises \(17-24,\) find the particular solution of the differential equation that satisfies the initial condition. $$ \text{Differential Equation} \quad \text{Initial Condition} $$ $$ y^{\prime}+(2 x-1) y=0 \quad y(1)=2 $$

Short answer

The particular solution of the given differential equation that satisfies the initial condition \(y(1)=2\) is \(y=2e e^{-x^2+x}\)

Step-by-step solution

Convert to Standard Form

The given differential equation is \(y^{\prime}+(2x-1)y = 0\). We can rewrite this in standard form as \(y^{\prime} + p(x)y = g(x)\), where \(p(x) = 2x-1\) and \(g(x)=0\). So the standard form is \(y^{\prime} + (2x-1)y = 0\).

Solve The Differential Equation

The solution for this form of differential equation is \(y= e^{-\int p(x)dx} \[ \int g(x)e^{\int p(x) dx} dx + C \]\). Given that \(g(x)=0\), the integral part of the equation will vanish, so the solution can be simplified to \(y= C e^{-\int p(x) dx}\). Substituting \(p(x) = 2x-1\) gives \(y= C e^{-\int (2x-1)dx}\). Thus, the general solution is \(y = C e^{-x^2+x}\).

Use initial Condition to find particular solution

According to the initial condition, when \(x=1\), \(y=2\). Substituting these values into the equation gives \(2= C e^{-(1)^2+(1)}\). Solving for \(C\) yields \(C=2e\). Thus, the particular solution satisfying the initial condition \(y(1)=2\) is \(y=2e e^{-x^2+x}\).

Key concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They serve as a powerful tool in modeling various physical systems where the rate of change of one variable is dependent on the variable itself. Consider the given exercise, we have a first-order linear differential equation of the form y' + (2x - 1)y = 0. This equation tells us that the rate of change of y (denoted by y') is related to y itself, modulated by the expression (2x - 1). Understanding how to manipulate and solve these equations is critical as they appear extensively in fields such as physics, engineering, economics, and biology. In our example, solving such an equation involves finding a function y that satisfies the equation for all values of x.

Initial Value Problem

An initial value problem in the context of differential equations is a problem where you are given a differential equation along with initial conditions, which are values of the unknown function and often its derivatives at a specific point. These conditions are used to find a unique solution to the differential equation that passes through the given point. In the case of our exercise, the initial condition provided is y(1) = 2. It specifies the value of the function y when x is 1. The goal is to use this piece of information to solve the differential equation for the specific case, effectively finding a function y that aligns with both the equation and the initial condition. This narrows down the solutions from a general solution typically involving an arbitrary constant C, to a particular solution that satisfies the given condition.

Integration Techniques

Integration techniques are methods used to perform integration, which is the mathematical process of finding the function given its derivative. This process is essentially the reverse of differentiation. In differential equations, integration is often used to find the general solution to an equation. The problem at hand requires integrating the function p(x) = 2x - 1 to find the solution to the differential equation. Common integration techniques include methods like substitution, integration by parts, partial fractions, and others, depending on the complexity of the function to be integrated. For the given exercise, the integration results in an exponential function involving x, which is then used along with the initial condition to solve for the constant C and find the particular solution of the given differential equation.