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Magazine Chapter 10: Problem 21
Solve the initial-value problem. $$y^{\prime}+2 y=1, y(0)=1$$
Short Answer
Expert verified
y = \frac{1}{2} (1 + e^{-2x}).
Step by step solution
01
Identify the form of the differential equation
The given differential equation is of the form \[ y^{\text{'} } + P(x) y = Q(x) \] where \(P(x)\) is 2 and \(Q(x)\) is 1.
02
Find the integrating factor
The integrating factor \( \text{I.F.} \) is given by \[ \text{I.F.} = e^{\tiny \text{ }\tiny \text{\tiny $\tiny {\tiny\boldintdot{\tiny P(x)} }}} = e^{\tiny 2x}. \]
03
Multiply through by the integrating factor
Multiplying the entire differential equation by \( e^{2x} \) gives:\[ e^{2x} y^{\text{'}} + 2 e^{2x} y = e^{2x}. \]
04
Simplify the left-hand side
The left-hand side of the equation is the derivative of \[ y e^{2x}. \] Therefore, the equation becomes: \[ \frac{d}{dx}(y e^{2x}) = e^{2x}. \]
05
Integrate both sides with respect to x
Integrate:\[ \tiny { \boldintdot{\tiny \frac{d}{dx}(y e^{2x}) \tiny \tiny d x} }= \tiny { \boldintdot {\tiny e^{2x} \tiny d x} }. \] The left-hand side integrates to \( y e^{2x}\), and the right-hand side integrates to \( \frac{1}{2} e^{2x} + C. \) Therefore we have:\[ y e^{2x} = \frac{1}{2} e^{2x} + C. \]
06
Solve for y
Divide by \( e^{2x} \):\[ y = \frac{1}{2} + C e^{-2x}. \]
07
Apply the initial condition
Use the initial condition \( y(0) = 1 \) to solve for C:\[ 1 = \frac{1}{2} + C \times e^{0}. \] Simplifying this, we get: \[ 1 = \frac{1}{2} + C \Rightarrow C = \frac{1}{2}. \]
08
Write the final solution
Substitute the value of \( C \) into the general solution to get the particular solution:\[ y = \frac{1}{2} + \frac{1}{2} e^{-2x}. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Differential Equations
A differential equation involves an unknown function and its derivatives. The goal is usually to find this unknown function. In our example, the given differential equation is: \[ y' + 2y = 1 \] Here, \( y' \) is the derivative of \( y \) with respect to \( x \), and the equation includes the function \( y \) and its first derivative. Differential equations are powerful tools in math and science, used to model various phenomena such as growth rates and motion. Break the problem down into steps:
- Identify the type of differential equation you're dealing with
- Find the solution step by step
The Integrating Factor Method
To solve linear first-order differential equations like ours, we use an integrating factor (I.F.). This method simplifies the equation, making it easier to solve. The general form of a linear first-order differential equation is: \[ y' + P(x) y = Q(x) \] For our equation \( y' + 2y = 1 \), we identify:
- \( P(x) = 2 \)
- \( Q(x) = 1 \)
Applying the Initial Condition
Once you've found the general solution to the differential equation, it's time to apply the initial condition. This is given as \( y(0) = 1 \) in our problem. From the integrating factor process, we had: \( y e^{2x} = \frac{1}{2} e^{2x} + C \) We solve for \( y \) by dividing through by \( e^{2x} \), resulting in: \( y = \frac{1}{2} + C e^{-2x} \). Next, use the initial condition (where \( x = 0 \)). Substituting \( y(0) = 1 \) gives us: \( 1 = \frac{1}{2} + C \times e^{0} \). Since \( e^{0} = 1 \), the equation simplifies to: \( 1 = \frac{1}{2} + C \). Solving for \( C \), we get: \( C = \frac{1}{2} \). Finally, substitute \( C \) back into the solution to obtain the particular solution: \( y = \frac{1}{2} + \frac{1}{2} e^{-2x} \). This completes the solution to our initial-value problem.
