Chapter 9: Problem 7
Solve the differential equations. \(2 y^{\prime}=e^{x / 2}+y\)
Short Answer
Expert verified
The solution is \(y = e^{\frac{1}{2}x} \left( \frac{x}{2} + C \right)\).
Step by step solution
01
Rewrite the Differential Equation
Begin by rearranging terms in the given differential equation \(2 y^{\prime} = e^{x / 2} + y\). Move the term involving \(y\) to the left side to isolate \(y^{\prime}\) as follows: \(2 y^{\prime} - y = e^{x / 2}\). Then rewrite it in standard linear first-order form: \(y^{\prime} - \frac{1}{2}y = \frac{1}{2} e^{x / 2}\).
02
Identify Integrating Factor
For a linear differential equation of the form \(y^{\prime} + P(x) y = Q(x)\), the integrating factor \(\mu(x)\) is given by \(e^{\int P(x) \, dx}\). Here, \(P(x) = -\frac{1}{2}\), so the integrating factor is: \[ \mu(x) = e^{-\int \frac{1}{2} \, dx} = e^{-\frac{1}{2}x} \].
03
Multiply the Differential Equation by the Integrating Factor
Multiply the entire differential equation by \(\mu(x) = e^{-\frac{1}{2}x}\). You get:\[e^{-\frac{1}{2}x} y^{\prime} - \frac{1}{2} e^{-\frac{1}{2}x} y = \frac{1}{2} e^{0} \].Notice that the left-hand side is the derivative of a product: \((e^{-\frac{1}{2}x} y)^{\prime}\).
04
Integrate Both Sides
Observe that \((e^{-\frac{1}{2}x} y)^{\prime} = \frac{1}{2}\). Integrate both sides with respect to \(x\):\(\int (e^{-\frac{1}{2}x} y)^{\prime} \, dx = \int \frac{1}{2} \, dx\), leading to:\[e^{-\frac{1}{2}x} y = \frac{x}{2} + C\], where \(C\) is the integration constant.
05
Solve for y
To find \(y\), multiply through by \(e^{\frac{1}{2}x}\) to get:\[y = e^{\frac{1}{2}x} \left( \frac{x}{2} + C \right)\].This represents the general solution for the differential equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integrating Factor
An integrating factor is a crucial tool for solving first-order linear differential equations. It is used to transform a non-exact differential equation into an exact one, enabling easier integration. The goal is to simplify the equation into a form where the derivative of a product appears, allowing for straightforward integration.
To find the integrating factor, consider a general first-order linear differential equation:
To find the integrating factor, consider a general first-order linear differential equation:
- \( y^{\prime} + P(x) y = Q(x) \)
General Solution
Finding the general solution to a differential equation involves determining a function or a set of functions that satisfy the equation. After applying the integrating factor, the equation is now in a form:\[(e^{-\frac{1}{2}x} y)^{\prime} = \frac{1}{2}\]The left side of the equation appears as a derivative of a product, which is a key indicator that integration can be performed straightforwardly. Integrating both sides gives:\[e^{-\frac{1}{2}x} y = \frac{x}{2} + C\]where \( C \) is an integration constant. This process transforms the differential equation into the general solution form. To express \( y \) explicitly, simply solve for \( y \) by multiplying both sides by \( e^{\frac{1}{2}x} \):
- \( y = e^{\frac{1}{2}x} \left( \frac{x}{2} + C \right) \)
Integration Constant
The integration constant, represented as \( C \) in the general solution, is a vital addition when integrating both sides of a differential equation. When integrating, the constant accounts for all possible initial conditions and scenarios that may be applied to the general solution.
Every differential equation represents a family of solutions, where each specific solution within that family corresponds to a particular initial condition or integration constant. For example, in the general solution:\[ y = e^{\frac{1}{2}x} \left( \frac{x}{2} + C \right)\]the integration constant \( C \) allows this equation to adapt to different initial values by providing a unique result for each value of \( C \).
Determining \( C \) often depends on additional information, such as initial conditions provided within a specific problem setting.
Every differential equation represents a family of solutions, where each specific solution within that family corresponds to a particular initial condition or integration constant. For example, in the general solution:\[ y = e^{\frac{1}{2}x} \left( \frac{x}{2} + C \right)\]the integration constant \( C \) allows this equation to adapt to different initial values by providing a unique result for each value of \( C \).
Determining \( C \) often depends on additional information, such as initial conditions provided within a specific problem setting.