Problem 19
Question
For \(y^{\prime}+\frac{x+2}{x+1} y=\frac{2 x e^{-x}}{x+1}\) an integrating factor is \(e^{\int[(x+2) /(x+1)] d x}=(x+1) e^{x},\) so \(\frac{d}{d x}\left[(x+1) e^{x} y\right]=2 x\) and \(y=\frac{x^{2}}{x+1} e^{-x}+\frac{c}{x+1} e^{-x}\) for \(-1< x<\infty .\) The entire solution is transient.
Step-by-Step Solution
Verified Answer
\( y = \frac{x^2}{x+1} e^{-x} + \frac{C}{x+1} e^{-x} \) for \(-1 < x < \infty\).
1Step 1: Identify the Given Equation
The given first-order linear differential equation is \(y^{\prime} + \frac{x+2}{x+1} y = \frac{2x e^{-x}}{x+1}\). We need to solve for \( y \) using an integrating factor.
2Step 2: Compute the Integrating Factor
We are given an integrating factor which is \( e^{\int \left(\frac{x+2}{x+1}\right) \,dx} = (x+1)e^x \). This integrating factor will be used to make the differential equation exact.
3Step 3: Apply the Integrating Factor
Multiply the entire equation by the integrating factor \((x+1)e^x\), converting it into \(\frac{d}{dx}[(x+1)e^x y] = 2x\). This transformation simplifies integrating the left side directly.
4Step 4: Integrate the Transformed Equation
Integrate the equation \(\frac{d}{dx}[(x+1)e^x y] = 2x\). The left side directly integrates to \((x+1)e^x y\), and for the right side, we integrate \(2x\) with respect to \(x\), resulting in \(x^2 + C\), where \(C\) is the constant of integration.
5Step 5: Solve for y
After integration, we have \((x+1)e^x y = x^2 + C\). Solve for \(y\) by dividing both sides by \((x+1)e^x\). Thus, \(y = \frac{x^2}{x+1} e^{-x} + \frac{C}{x+1} e^{-x}\).
6Step 6: Conclusion
The complete solution is transient for \(-1 < x < \infty\), where transient refers to the behavior of solutions as they approach zero. The complete solution demonstrates how constants affect general solutions.
Key Concepts
Integrating Factor MethodExact Differential EquationTransient SolutionIntegration Techniques
Integrating Factor Method
The integrating factor method is a very useful technique for solving linear first-order differential equations. This method helps in transforming a non-exact equation into an exact one by multiplying it with a strategically chosen function. The purpose of this special function, called the integrating factor, is to allow the differential equation to be written in a simpler form, making it amenable to straightforward integration.
To find the integrating factor for a differential equation of the form \(y' + p(x)y = q(x)\), you calculate \(e^{\int p(x) \, dx}\). This exponential function adjusts the equation so that the left-hand side becomes the derivative of a product, specifically \(\frac{d}{dx}[\text{{Integrating Factor}} \cdot y]\).
To find the integrating factor for a differential equation of the form \(y' + p(x)y = q(x)\), you calculate \(e^{\int p(x) \, dx}\). This exponential function adjusts the equation so that the left-hand side becomes the derivative of a product, specifically \(\frac{d}{dx}[\text{{Integrating Factor}} \cdot y]\).
- Identify \(p(x)\) from the equation.
- Compute the integral of \(p(x)\).
- Take the exponential of the integral.
Exact Differential Equation
An exact differential equation is one that can be expressed in the form of a total differential. This means the equation can be seen as the derivative of some function. Once the integrating factor method is applied, the non-exact equation we started with becomes exact.
For an equation like \(M(x, y)\,dx + N(x, y)\,dy = 0\), it is exact if \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\). However, if the equation is not exact, introducing an integrating factor often adjusts it to this form.
For an equation like \(M(x, y)\,dx + N(x, y)\,dy = 0\), it is exact if \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\). However, if the equation is not exact, introducing an integrating factor often adjusts it to this form.
- Exact equations simplify the solving process.
- After finding an appropriate integrating factor, integration becomes direct.
Transient Solution
The term transient solution in the context of differential equations refers to the part of the solution that fades or vanishes over time, approaching zero as time goes on or as \(x\to\infty\).
In our specific solution \(y = \frac{x^2}{x+1} e^{-x} + \frac{C}{x+1} e^{-x}\), the entire expression is transient. This is because each part involves an exponential decay term \(e^{-x}\). This exponential factor ensures that, as \(x\) increases, the function values decrease and tend towards zero.
In our specific solution \(y = \frac{x^2}{x+1} e^{-x} + \frac{C}{x+1} e^{-x}\), the entire expression is transient. This is because each part involves an exponential decay term \(e^{-x}\). This exponential factor ensures that, as \(x\) increases, the function values decrease and tend towards zero.
- Transient solutions model temporary phenomena.
- They differ from steady-state solutions which remain constant.
Integration Techniques
Integration techniques are crucial for solving differential equations once they have been transformed into an integrable form by a suitable method, such as the integrating factor method. Recognizing when and how to integrate these equations is key.
For example, in the exercise at hand, the transformation allows the left-hand side to simplify to \(\frac{d}{dx}[(x+1)e^x y]\), which directly integrates to \((x+1)e^x y\).
On the right side, you integrate the polynomial expression \(2x\) with respect to \(x\) to get \(x^2\), plus a constant of integration \(C\).
For example, in the exercise at hand, the transformation allows the left-hand side to simplify to \(\frac{d}{dx}[(x+1)e^x y]\), which directly integrates to \((x+1)e^x y\).
On the right side, you integrate the polynomial expression \(2x\) with respect to \(x\) to get \(x^2\), plus a constant of integration \(C\).
- Techniques like substitution or partial fractions often assist in integration.
- Understanding the type of function you are integrating improves accuracy.
- Recognize common integrals to streamline solving processes.
Other exercises in this chapter
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