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Magazine Chapter 31: Problem 34
Find the indicated particular solutions of the given differential equations. \(y^{\prime} \sqrt{x}+\frac{1}{2} y=e^{\sqrt{x}} ; \quad x=1\) when \(y=3\)
Short Answer
Expert verified
The particular solution is: \( y = \frac{1}{2} e^{\sqrt{x}} + \frac{3e - \frac{1}{2} e^{2}}{e^{\sqrt{x}}} \).
Step by step solution
01
Identify the Type of Differential Equation
The given differential equation is \( y^{\prime} \sqrt{x} + \frac{1}{2} y = e^{\sqrt{x}} \). This is a first-order linear differential equation since it can be written in the form \( y^{\prime} + P(x) y = Q(x) \). The equation needs to be transformed or solved using an appropriate method for first-order linear equations.
02
Rewrite the Equation in Standard Form
Rewrite the differential equation in the standard form by dividing the entire equation by \( \sqrt{x} \): \[y^{\prime} + \frac{1}{2\sqrt{x}} y = \frac{e^{\sqrt{x}}}{\sqrt{x}}\]Thus, the equation is now in the form \( y^{\prime} + P(x)y = Q(x) \), where \( P(x) = \frac{1}{2\sqrt{x}} \) and \( Q(x) = \frac{e^{\sqrt{x}}}{\sqrt{x}} \).
03
Find the Integrating Factor
The integrating factor \( \mu(x) \) is found by calculating \( e^{\int P(x) \, dx} \). We have:\[\int P(x) \, dx = \int \frac{1}{2\sqrt{x}} \, dx = \int \frac{1}{2} x^{-1/2} \, dx = x^{1/2} = \sqrt{x}\]Therefore, the integrating factor is \( \mu(x) = e^{\sqrt{x}} \).
04
Multiply Through by the Integrating Factor
Multiply the standard form of the equation by the integrating factor \( \mu(x) = e^{\sqrt{x}} \):\[e^{\sqrt{x}} y^{\prime} + e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} y = e^{\sqrt{x}} \cdot \frac{e^{\sqrt{x}}}{\sqrt{x}}\]Simplify the right-hand side:\[e^{\sqrt{x}} y^{\prime} + \frac{1}{2} y e^{\sqrt{x}} = \frac{e^{2\sqrt{x}}}{\sqrt{x}}\]
05
Recognize the Left-Hand Side as a Derivative
The left-hand side can now be recognized as the derivative of a product, because it follows the form of \[\frac{d}{dx} \left(e^{\sqrt{x}} y \right)\]Therefore,\[\frac{d}{dx} \left(e^{\sqrt{x}} y \right) = \frac{e^{2\sqrt{x}}}{\sqrt{x}}\]
06
Integrate Both Sides
Integrate both sides with respect to \( x \):\[\int \frac{d}{dx} \left(e^{\sqrt{x}} y \right) \, dx = \int \frac{e^{2\sqrt{x}}}{\sqrt{x}} \, dx\]On the left-hand side, we have:\[e^{\sqrt{x}} y = \int \frac{e^{2\sqrt{x}}}{\sqrt{x}} \, dx + C\]We'll solve the right-hand side later, but for now, note the result from integrating the derivative expression.
07
Evaluate the Integral
Evaluating \( \int \frac{e^{2\sqrt{x}}}{\sqrt{x}} \, dx \) is complex and may require substitution. Let \( u = 2\sqrt{x} \), hence \( du = \frac{1}{\sqrt{x}} \, dx \), so the integral becomes:\[\int e^{u} \, \frac{1}{2} \, du = \frac{1}{2} e^{u} + C = \frac{1}{2} e^{2\sqrt{x}} + C\]
08
Return to Solve for y and Use Initial Condition
Substitute back: \[e^{\sqrt{x}} y = \frac{1}{2} e^{2\sqrt{x}} + C\]At \( x = 1 \), \( y = 3 \):\[e^{1} \cdot 3 = \frac{1}{2} e^{2} + C\]\[3e = \frac{1}{2} e^{2} + C \quad \Rightarrow \quad C = 3e - \frac{1}{2} e^{2}\]
09
Write the Particular Solution
Substitute \( C \) back into the equation:\[y = \frac{1}{2} e^{\sqrt{x}} + \frac{3e - \frac{1}{2} e^{2}}{e^{\sqrt{x}}}\]This represents the particular solution for the differential equation with initial condition \( y(1) = 3 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integrating Factor
In solving first-order linear differential equations, the integrating factor is a crucial concept. Think of it as a helpful tool that transforms the equation into an easily integrable form. The integrating factor is a function, usually denoted as \( \mu(x) \), designed to simplify the differential equation. Here's how it works:
1. For a standard form differential equation \( y' + P(x)y = Q(x) \), the integrating factor is given by \( \mu(x) = e^{\int P(x) \, dx} \).
2. Multiplying the entire differential equation by this integrating factor turns the left side into the derivative of a product. This reduction makes the equation simpler to solve.
3. In our example, for \( P(x) = \frac{1}{2\sqrt{x}} \), we found the integrating factor to be \( e^{\sqrt{x}} \), which transformed the equation nicely.
By using the integrating factor, solving for \( y \) involves straightforward integration rather than more complex differential equation techniques. It's like using a key to unlock a lock, making the process much easier.
1. For a standard form differential equation \( y' + P(x)y = Q(x) \), the integrating factor is given by \( \mu(x) = e^{\int P(x) \, dx} \).
2. Multiplying the entire differential equation by this integrating factor turns the left side into the derivative of a product. This reduction makes the equation simpler to solve.
3. In our example, for \( P(x) = \frac{1}{2\sqrt{x}} \), we found the integrating factor to be \( e^{\sqrt{x}} \), which transformed the equation nicely.
By using the integrating factor, solving for \( y \) involves straightforward integration rather than more complex differential equation techniques. It's like using a key to unlock a lock, making the process much easier.
Initial Conditions
Initial conditions are essential in determining a specific solution out of many potential solutions to a differential equation. An initial condition provides a value at a certain point, which the solution curve must pass through. Here's how it is applied:
- Suppose you're given a differential equation and the initial condition \( y(1) = 3 \).
- This means that when the input variable \( x \) is 1, the output, \( y \), must be 3.
- In the solving process, such conditions help us find the constant of integration (denoted \( C \)). This ensures that the resulting function will cross the specified point, tailoring the general solution to a particular one.
- For example, after finding the general form of the solution \( y \), you substitute \( x = 1 \) and \( y = 3 \) into the equation to solve for \( C \).
Initial conditions make differential equations highly practical, allowing us to find specific solutions that model real-world problems accurately.
- Suppose you're given a differential equation and the initial condition \( y(1) = 3 \).
- This means that when the input variable \( x \) is 1, the output, \( y \), must be 3.
- In the solving process, such conditions help us find the constant of integration (denoted \( C \)). This ensures that the resulting function will cross the specified point, tailoring the general solution to a particular one.
- For example, after finding the general form of the solution \( y \), you substitute \( x = 1 \) and \( y = 3 \) into the equation to solve for \( C \).
Initial conditions make differential equations highly practical, allowing us to find specific solutions that model real-world problems accurately.
Standard Form Differential Equation
Rewriting a differential equation in its standard form is often the first step in the solving process, especially for first-order linear differential equations. Let's explore why this is important:
- The standard form for these equations is \( y' + P(x)y = Q(x) \). Placing an equation in this form guides us toward appropriate solution strategies, such as the integrating factor method.
- In the exercise, the original equation was \( y^{\prime} \sqrt{x} + \frac{1}{2} y = e^{\sqrt{x}} \). By dividing through by \( \sqrt{x} \), we obtained a standard form making it easier to identify \( P(x) \) and \( Q(x) \).
- The transformation into standard form simplifies the recognition of how the variable \( y \) interacts with \( x \), allowing us to systematically solve the differential equation.
Getting the equation into standard form is like setting a solid foundation in mathematics, ensuring the method of solution works smoothly, leading us to a clear and precise result.
- The standard form for these equations is \( y' + P(x)y = Q(x) \). Placing an equation in this form guides us toward appropriate solution strategies, such as the integrating factor method.
- In the exercise, the original equation was \( y^{\prime} \sqrt{x} + \frac{1}{2} y = e^{\sqrt{x}} \). By dividing through by \( \sqrt{x} \), we obtained a standard form making it easier to identify \( P(x) \) and \( Q(x) \).
- The transformation into standard form simplifies the recognition of how the variable \( y \) interacts with \( x \), allowing us to systematically solve the differential equation.
Getting the equation into standard form is like setting a solid foundation in mathematics, ensuring the method of solution works smoothly, leading us to a clear and precise result.
