Problem 44 - Calculus and its applications

Question

Solve for \(y .\) \(y^{\prime}=2 x-x y ; \quad y=9\) when \(x=0\)

Step-by-Step Solution

Answer

The solution is \( y = -2 + 11e^{\frac{x^2}{2}} \).

Step 1: Recognize the Type of Differential Equation

The given differential equation is \( y' = 2x - xy \). This is a first-order linear differential equation in the form \( y' + P(x)y = Q(x) \), where \( P(x) = -x \) and \( Q(x) = 2x \).

Step 2: Find the Integrating Factor

The integrating factor \( \mu(x) \) for a linear differential equation can be calculated using the formula \( \mu(x) = e^{\int P(x) \, dx} \). Here, \( P(x) = -x \), so we integrate: \( \int -x \, dx = -\frac{x^2}{2} \). Thus, \( \mu(x) = e^{-\frac{x^2}{2}} \).

Step 3: Multiply the Equation by the Integrating Factor

Multiply the entire differential equation \( y' + (-x)y = 2x \) by the integrating factor \( \mu(x) = e^{-\frac{x^2}{2}} \) to obtain: \[ e^{-\frac{x^2}{2}}y' + e^{-\frac{x^2}{2}}(-x)y = e^{-\frac{x^2}{2}}2x. \]

Step 4: Recognize the Left Side as a Derivative

Notice that the left side of the equation \( e^{-\frac{x^2}{2}}y' + e^{-\frac{x^2}{2}}(-x)y \) can be rewritten as the derivative of the product of the integrating factor and \( y \). This can be expressed as: \[ \frac{d}{dx}(e^{-\frac{x^2}{2}}y) = e^{-\frac{x^2}{2}}2x. \]

Step 5: Integrate Both Sides

Integrate both sides with respect to \( x \):\[ \int \frac{d}{dx}(e^{-\frac{x^2}{2}}y) \, dx = \int e^{-\frac{x^2}{2}}2x \, dx. \]The left side simplifies to \( e^{-\frac{x^2}{2}}y \), and for the right side, we can use substitution to find the integral.

Step 6: Simplify the Right-side Integral

Let \( u = -\frac{x^2}{2} \), then \( du = -x\,dx \). So, \( \int e^{u} (-2\,du) = -2 \int e^{u} \, du \). This integral becomes \( -2e^{u} + C = -2e^{-\frac{x^2}{2}} + C \).

Step 7: Solve for y

The expression becomes: \[ e^{-\frac{x^2}{2}}y = -2e^{-\frac{x^2}{2}} + C. \] Multiply through by \( e^{\frac{x^2}{2}} \) to solve for \( y \):\[ y = -2 + Ce^{\frac{x^2}{2}}. \]

Step 8: Apply Initial Conditions

Given that \( y = 9 \) when \( x = 0 \), substitute \( x = 0 \) and \( y = 9 \) into the equation: \[ 9 = -2 + C \cdot e^{0}. \]Solving for \( C \), we get \( C = 11 \).

Step 9: State the Final Solution

Substitute \( C = 11 \) back into the expression for \( y \):\[ y = -2 + 11e^{\frac{x^2}{2}}. \] This is the solution to the differential equation given the initial conditions.

Key Concepts

Initial Value Problem

An initial value problem (IVP) is a type of differential equation that comes with specific initial conditions. It's like being given a starting point on a road trip and figuring out which path to take to reach the destination. In this exercise, you start with a differential equation and an initial condition given as the value of the function at a certain point. Here, you have the equation \( y' = 2x - xy \) and an initial condition where \( y = 9 \) when \( x = 0 \).

• The goal is to find a function \( y(x) \) that satisfies both the differential equation and the initial condition.
• This initial condition helps to determine the unique path or solution corresponding to the particular scenario described by the equation.

IVPs are fundamental in the study of differential equations as they reflect real-world scenarios where the state at the initial time influences the future outcomes.

Integrating Factor

The integrating factor is a clever mathematical tool used to solve first-order linear differential equations. Imagine it as a magical multiplier that simplifies the equation. Here, we need to turn the given differential equation into a form that's easier to integrate. To do this, we calculate the integrating factor \( \mu(x) = e^{\int P(x) \, dx} \), where \( P(x) \) is part of the equation.

• For the given problem, \( P(x) = -x \), so we compute \( \mu(x) = e^{-\frac{x^2}{2}} \).
• This function is multiplied throughout the equation, transforming it into a simpler expression that reveals hidden structures.

Integrating factors are a key technique in taming complex differential equations, allowing solutions to be found in a straightforward manner.

First-order Linear Differential Equation

A first-order linear differential equation is an equation that involves the first derivative of a function, and it can be expressed in the standard form \( y' + P(x)y = Q(x) \). In our example, the equation \( y' = 2x - xy \) fits this format, where:

• \( P(x) = -x \)
• \( Q(x) = 2x \)

These equations commonly appear in modeling situations that involve rates of change, such as population growth or radioactive decay. By identifying the form as linear, we can apply standard methods like the integrating factor technique to solve them. The beauty of first-order linear equations lies in their predictability and the systematic approach they offer to finding solutions.

Solution of Differential Equations

The solution to a differential equation is a function that satisfies the original equation over a particular range. For the problem at hand, solving the differential equation with the given initial value involves several steps:

• First, recognize the form of the equation and apply the integrating factor \( \mu(x) = e^{-\frac{x^2}{2}} \).
• Next, transform the original equation to its simplest form by multiplying it with the integrating factor.
• This leads to recognizing the left side of the modified equation as the derivative of a product, which makes integration straightforward.
• Finally, integrate both sides and apply the initial condition to solve for the constant of integration, \( C \).

Ultimately, we find that \( y = -2 + 11e^{\frac{x^2}{2}} \) is the solution, satisfying both the differential equation and the initial condition. This approach highlights how mathematical methods bring clarity and structure to problems that at first seem complex.