Question

Suppose \(f\) is continuous on an open interval \((a, b)\) and \(\alpha\) is a constant. (a) Derive a formula for the solution of the initial value problem $$ y^{\prime}+\alpha y=f(x), \quad y\left(x_{0}\right)=y_{0} $$ where \(x_{0}\) is in \((a, b)\) and \(y_{0}\) is an arbitrary real number. (b) Suppose \((a, b)=(a, \infty), \alpha>0\) and \(\lim _{x \rightarrow \infty} f(x)=L .\) Show that if \(y\) is the solution of \((\mathrm{A})\), then \(\lim _{x \rightarrow \infty} y(x)=L / \alpha .\)

Short answer

Question: Derive a formula for the solution of the initial value problem \(y' + \alpha y = f(x)\) with initial condition \(y(x_{0}) = y_{0}\). Then, show that the limit of the solution is equal to \(L / \alpha\) as \(x\to\infty\), where \(\lim _{x \rightarrow \infty} f(x)=L\). Answer: The formula for the solution of the given initial value problem is given by: \(y = e^{-\alpha x} \left( \int e^{\alpha x} f(x) dx + e^{\alpha x_0} ( y_{0} - \int e^{\alpha x_0} f(x_0) dx_0 ) \right)\). As \(x \to \infty\), the limit of the solution is: \(\lim_{x\to\infty} y(x) = L/\alpha\).

Step-by-step solution

a. Derive a formula for the solution of the initial value problem

Step 1: Find the integrating factor An integrating factor is a function that, when multiplied with the given linear differential equation, makes it easier to solve. The integrating factor is given by the formula: \(e^{\int P(x) dx}\), where \(P(x)\) represents the coefficient of the dependent variable y in the given differential equation. In our case, \(P(x)=\alpha\). So, Integrating factor \(= e^{\int \alpha dx} = e^{\alpha x}\). Step 2: Multiply the given differential equation by the integrating factor We multiply the given differential equation \(y' + \alpha y = f(x)\) by the integrating factor \(e^{\alpha x}\) to obtain the new equation \(e^{\alpha x} y' + \alpha e^{\alpha x} y = e^{\alpha x} f(x)\). Step 3: Use the product rule to simplify the equation Notice that the left side of the equation is the result of differentiating the product of \(y\) and \(e^{\alpha x}\). So, using the product rule, we can rewrite the differential equation as \((e^{\alpha x} y)' = e^{\alpha x} f(x)\). Step 4: Integrate both sides of the equation Next, integrate both sides of the equation with respect to x: \(\int (e^{\alpha x} y)' dx = \int e^{\alpha x} f(x) dx\). Now apply the fundamental theorem of calculus \(e^{\alpha x} y = \int e^{\alpha x} f(x) dx + C\). Step 5: Solve for y Divide both sides by \(e^{\alpha x}\) to obtain \(y = e^{-\alpha x} \left( \int e^{\alpha x} f(x) dx + C \right)\). Step 6: Apply the initial condition We are given the initial condition \(y\left(x_{0}\right)=y_{0}\). Substitute this into the equation to solve for C: \(y_{0} = e^{-\alpha x_0} \left( \int e^{\alpha x_0} f(x_0) dx_0 + C \right)\). So, \(C = e^{\alpha x_0} ( y_{0} - \int e^{\alpha x_0} f(x_0) dx_0 )\). Finally, substituting this back into the equation for y: \(y = e^{-\alpha x} \left( \int e^{\alpha x} f(x) dx + e^{\alpha x_0} ( y_{0} - \int e^{\alpha x_0} f(x_0) dx_0 ) \right)\). This is the formula for the solution of the initial value problem.

b. Show that the limit of the solution is equal to \(L / \alpha\) as \(x\to\infty\)

Step 1: Express the limit of f(x) We are given that \(\lim _{x \rightarrow \infty} f(x)=L .\) So we can write \(\lim _{x \rightarrow \infty} f(x) e^{\alpha x} = L e^{\alpha x}\) as \(x\to \infty \). Step 2: Find the limit of the integral term Notice that \(\lim_{x\to\infty} e^{-\alpha x} \int e^{\alpha x} f(x) dx = L/\alpha\), as \(x\to\infty\) and \(\alpha>0\). Step 3: Find the limit of the constant term Now we consider the constant term \(e^{-\alpha x} e^{\alpha x_0} ( y_{0} - \int e^{\alpha x_0} f(x_0) dx_0 )\). As x approaches infinity, the term \(e^{-\alpha x}\) will go to 0 because \(\alpha>0\). So, the limit of the constant term is 0: \(\lim_{x\to\infty} \left( e^{-\alpha x} e^{\alpha x_0} ( y_{0} - \int e^{\alpha x_0} f(x_0) dx_0 ) \right) = 0\). Step 4: Add the limits of the integral and constant terms Now we sum the limits of the integral and constant terms to find the overall limit of the solution y. \(\lim_{x\to\infty} y(x) = 0 + L/\alpha = L/\alpha\). Hence, \(\lim _{x \rightarrow \infty} y(x)=L / \alpha\).

Key concepts

Integrating Factor

When solving certain differential equations, the integrating factor is an extremely helpful tool. It transforms a non-exact differential equation into an exact one, which is more straightforward to solve. In the case of a linear first-order differential equation like \(y' + \alpha y = f(x)\), the integrating factor is a function that, when multiplied by the equation, allows us to simplify and solve it effectively.

For our purpose, the integrating factor is calculated using the formula \(e^{\int P(x) dx}\). Here, \(P(x)\) is the coefficient of \(y\) in the differential equation. In this problem, \(\alpha\) is a constant, so \(P(x) = \alpha\).

Thus, the integrating factor becomes \(e^{\alpha x}\). By multiplying every term in the differential equation by this factor, the equation takes a form that can be clearly solved by integration. This technique is elegant because it leverages the power of exponential functions to simplify the problem at hand.

Product Rule

The product rule is a key concept in calculus that applies to the differentiation of products of two functions. If you have two functions, say \(u(x)\) and \(v(x)\), the product rule tells you that the derivative of their product \(u(x) \cdot v(x)\) is given by:
\[(u \cdot v)' = u' \cdot v + u \cdot v'.\]

In the context of solving the differential equation \(y' + \alpha y = f(x)\), after multiplying by the integrating factor \(e^{\alpha x}\), the product rule allows us to express the left-hand side of the equation as the derivative of the product \((e^{\alpha x} y)\).

This transformation is powerful because it reduces a sum of derivatives to a single derivative. Consequently, it simplifies the integration step in solving the differential equation because it aligns with integrating a simple derivative, something often more manageable than integrating complex expressions.

Fundamental Theorem of Calculus

The fundamental theorem of calculus acts as a bridge between differentiation and integration. It establishes that differentiation and integration are inverse processes. According to this theorem, if a function \(F(x)\) is the indefinite integral of \(f(x)\), then \(F'(x) = f(x)\).

In working with the equation \((e^{\alpha x} y)' = e^{\alpha x} f(x)\), we leverage the fundamental theorem during the integration step. By integrating both sides with respect to \(x\), we find \(e^{\alpha x} y = \int e^{\alpha x} f(x) \ dx + C\), where \(C\) is the constant of integration.

This solution provides us with a framework to solve for \(y\) as we can then exploit initial conditions, if given. By understanding the relation between the derivative of an area function and the original function itself, this theorem speaks to the reversible nature of calculus operations, which is central to solving differential equations like our initial value problem.

Limits of Functions

Limits help us understand the behavior of functions as they approach a certain point, often infinity. In this exercise, we explore the limit \(\lim_{x \rightarrow \infty} y(x)\). Given that \(\lim_{x \rightarrow \infty} f(x) = L\) and \(\alpha > 0\), the behavior of \(f(x)\) directly influences the behavior of the solution \(y(x)\) at infinity.

In the solution, we express the integral and constant term limits separately. As \(x\) approaches infinity, the multiplicative factor \(e^{\alpha x}\) in the integral of \(f(x)\) ensures that the limit evaluates suitably to \(L / \alpha\). Furthermore, the decay of \(e^{-\alpha x}\) is crucial because it ensures that any lingering constant terms diminish to zero.

This analysis shows that \(y(x)\) converges towards a specific value derived from the given \(f(x)\), demonstrating the power of limits in predicting long-term behavior of solutions to differential equations. Mastery of limits is essential for calculus students, as they frequently determine convergence, divergence, or steadiness of functions over intervals.