Chapter 11: Problem 12
use separation of variables to find the solution to the differential equation subject to the initial condition. $$\frac{d I}{d x}=0.2 I, \quad I=6 \text { where } x=-1$$
Short Answer
Expert verified
The solution is \( I = 6e^{0.2(x+1)} \).
Step by step solution
01
Write Down the Differential Equation
The differential equation given is \( \frac{dI}{dx} = 0.2I \). This is a first-order linear differential equation.
02
Separate Variables
To solve by separation of variables, rearrange the equation to isolate the variables. Divide both sides by \( I \) and multiply both sides by \( dx \) to get \( \frac{1}{I} \, dI = 0.2 \, dx \).
03
Integrate Both Sides
Integrate \( \int \frac{1}{I} \, dI = \int 0.2 \, dx \). This gives \( \ln |I| = 0.2x + C \), where \( C \) is the integration constant.
04
Solve for the Constant Using Initial Condition
Use the initial condition \( I = 6 \) when \( x = -1 \) to find \( C \). Plug these values into the equation: \( \ln 6 = 0.2(-1) + C \). Solve for \( C \) to get \( C = \ln 6 + 0.2 \).
05
Solve for I
Rewrite the equation \( \ln |I| = 0.2x + \ln 6 + 0.2 \) to solve for \( I \). Exponentiate both sides to get \( |I| = e^{0.2x + \ln 6 + 0.2} \).
06
Simplify the Expression
Since \( I > 0 \) (as given initially \( I = 6 \)), remove the absolute value to get \( I = 6e^{0.2x + 0.2} = 6e^{0.2(x+1)} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Separation of Variables
Separation of Variables is a method used to solve differential equations, where we aim to separate the variables on different sides of the equation. It is particularly useful for first-order differential equations. The primary goal is to express the equation so that each variable, along with its differential, is isolated.
Consider the differential equation \( \frac{dI}{dx} = 0.2I \). To apply separation of variables, we rearrange terms to have differentials with the respective variables. In this case, divide both sides by \( I \) and multiply both sides by \( dx \), resulting in \( \frac{1}{I} \, dI = 0.2 \, dx \).
Consider the differential equation \( \frac{dI}{dx} = 0.2I \). To apply separation of variables, we rearrange terms to have differentials with the respective variables. In this case, divide both sides by \( I \) and multiply both sides by \( dx \), resulting in \( \frac{1}{I} \, dI = 0.2 \, dx \).
- This equation allows us to integrate each side independently with respect to its variable.
- It's crucial to maintain proper notation to avoid mixing variables and their differentials.
Initial Conditions
Initial Conditions provide the necessary data to find a unique solution for a differential equation. They specify the value of the unknown function at a particular point. These conditions help determine the integration constant.In the given differential equation \( \frac{dI}{dx} = 0.2I \), the initial condition is specified as \( I=6 \) when \( x=-1 \).
- This piece of information pinpoints the curve's exact path among potentially infinite solutions.
- Without an initial condition, the integration process leaves us with a family of curves, each represented by a different integration constant.
First-order Linear Differential Equation
First-order linear differential equations are characterized by having the form \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( P(x) \) and \( Q(x) \) are functions of \( x \) alone. The equation \( \frac{dI}{dx} = 0.2I \) is a straightforward example, where it can be rewritten in the form \( \frac{dy}{dx} - 0.2I = 0 \).These types of equations are called linear because the unknown function and its derivative appear in a linear fashion.
- The solution often involves the use of the integrating factor method or separation of variables, as in our example.
- Simplicity makes them a great starting point for students learning about differential equations.
Integration Constant
The Integration Constant, often represented by \( C \), is an arbitrary constant that appears when integrating an indefinite integral. Its role is crucial in providing a complete family of solutions for differential equations.After integrating a separated equation, as in \( \ln |I| = 0.2x + C \), \( C \) expresses that there are many solutions that differ by a constant value. This constant embodies the "lost" information during differentiation.
When initial conditions are supplied, like \( I=6 \) at \( x=-1 \), they serve to resolve the value of \( C \). It's fascinating how just one condition can specify the exact solution by pinning \( C \).
When initial conditions are supplied, like \( I=6 \) at \( x=-1 \), they serve to resolve the value of \( C \). It's fascinating how just one condition can specify the exact solution by pinning \( C \).
- In our example, substituting \( I = 6 \) and \( x = -1 \) determines that \( C = \ln 6 + 0.2 \).
- Specifying \( C \) converts a broad general solution into a precise particular solution, tailor-fit to the problem context.