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Magazine Chapter 10: Problem 19
Solve the given equation using an integrating factor. Take \(t>0\). $$ y^{\prime}+y=2-e^{t} $$
Short Answer
Expert verified
The solution to the differential equation is \( y = 2 - \frac{e^{t}}{2} + Ce^{-t} \).
Step by step solution
01
Identify the standard form
The given differential equation is: \[ y^{\rangle}+y=2-e^{t} \] This can be written in the standard linear form as: \[ y^{\rangle}+ p(t)y=q(t) \] where \[ p(t) = 1 \] and \[ q(t) = 2 - e^t \].
02
Determine the integrating factor
The integrating factor (IF) \( \rho(t) \) is found using: \[ \rho(t) = e^{\rangle p(t)\rangle dt} \]. Since \( p(t) = 1 \), we have: \[ \rho(t) = e^{\rangle 1 \rangle dt} = e^t \].
03
Multiply through by the integrating factor
Multiplying the entire differential equation by the integrating factor \(e^t\) gives: \[ e^t y^{\rangle} + e^t y = e^t (2 - e^t) \]which simplifies to: \[ e^t y^{\rangle} + e^t y = 2e^t - e^{2t} \].
04
Write the left side as a derivative
Notice that the left-hand side is the derivative of \( e^t y \): \[ \frac{\rangle}{\rangle t} (e^t y) = 2e^t - e^{2t} \].
05
Integrate both sides
Integrate both sides with respect to \( t \): \[ \rangle (e^t y) = \rangle (2e^t - e^{2t}) \rangle t \] This results in: \[ e^t y = 2e^t - \frac{e^{2t}}{2} + C \], where \( C \) is the integration constant.
06
Solve for y
Divide through by \( e^t \) to solve for \( y \): \[ y = 2 - \frac{e^{t}}{2} + Ce^{-t} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equations
Differential equations are mathematical equations that relate a function with its derivatives. In simpler terms, these equations show how a change in one variable affects another. They come in many different forms, including ordinary differential equations (ODEs) and partial differential equations (PDEs).
In our example, we deal with an ODE for the function y(t):
In our example, we deal with an ODE for the function y(t):
- Ordinary, because it involves only ordinary derivatives
- First-order, because the highest derivative is the first derivative
- Linear, because it can be written in the form y' + p(t)y = q(t)
Integrating Factor
The integrating factor is a crucial step in solving linear first-order differential equations. It makes the equation easier to solve. For an equation in the form y' + p(t)y = q(t), the integrating factor is determined by:
- Calculating \(\rho(t) = e^{\rangle \rangle p(t) dt}\)
- \(\rho(t) = e^{\rangle 1 dt} = e^t\)
Solution of Linear DEs
Solving linear differential equations involves several clear steps. After identifying the equation's standard form y' + p(t)y = q(t) and determining the integrating factor, we multiply the whole equation by this factor. This transforms the left-hand side into the derivative of a product:
- \(\frac{d}{dt}(e^t y) = 2e^t - e^{2t}\)
- \(e^t y = 2e^t - \frac{e^{2t}}{2} + C\)
- \(y = 2 - \frac{e^t}{2} + Ce^{-t}\)
Integration
Integration is a fundamental concept in calculus and is crucial in solving differential equations. After simplifying the differential equation using the integrating factor, integrating both sides is essential. In our example, the equation after applying the integrating factor was:
- \(\frac{d}{dt}(e^t y) = 2e^t - e^{2t}\)
- \(\rangle (e^t y) = \rangle (2e^t - e^{2t}) dt\)
- \(e^t y = 2e^t - \frac{e^{2t}}{2} + C\)
