Question

Consider the solutions \(Y(x)\) of $$ Y^{\prime}(x)+a Y(x)=d e^{-b x} $$ with \(a, b, d\) constants and \(a, b>0 .\) Calculate $$ \lim _{x \rightarrow \infty} Y(x) $$ Hint: \(\quad\) Consider separately the cases \(a \neq b\) and \(a=b\).

Short answer

\( \text{lim}_{x \rightarrow \infty} Y(x) = 0 \).

Step-by-step solution

Identify the Type of Differential Equation

The given differential equation is a first-order linear differential equation of the form: \( Y'(x) + aY(x) = d e^{-bx} \).

Solve the Homogeneous Equation

The homogeneous part of the equation is: \( Y'(x) + aY(x) = 0 \), which has the solution: \( Y_h(x) = Ce^{-ax} \) where \(C\) is the integration constant.

Find a Particular Solution

For the non-homogeneous equation, look for a particular solution of the form: \( Y_p(x) = A e^{-bx} \). Substitute \( Y_p(x) = A e^{-bx} \) into the original equation to find \(A\): \( -Ab e^{-bx} + aAe^{-bx} = de^{-bx} \). Simplify to get: \( A(a - b) = d \).

Solve for the Particular Solution Parameter

Case 1: \( a eq b \): \( A = \frac{d}{a-b} \). So, the particular solution is: \( Y_p(x) = \frac{d}{a-b} e^{-bx} \).

General Solution

The general solution to the differential equation is: \( Y(x) = Y_h(x) + Y_p(x) = Ce^{-ax} + \frac{d}{a-b} e^{-bx} \).

Analyze the Limit as x Approaches Infinity for a ≠ b

As \( x \rightarrow \infty \), both terms \( Ce^{-ax} \) and \( \frac{d}{a-b} e^{-bx} \) approach 0 because \( a, b > 0 \). Thus, \( \text{lim}_{x \rightarrow \infty} Y(x) = 0 \).

Consider Case When a = b

For \( a = b \), a different particular solution is required. Try a particular solution of the form \( Y_p(x) = Ax e^{-ax} \). Substitute it back into the differential equation and solve for \( A \): \( (-Aax + A)e^{-ax} + aAx e^{-ax} = de^{-ax} \). Simplifying to find \( A \), we get \( A = d \). Thus, the particular solution is \( Y_p(x) = dx e^{-ax} \).

General Solution for a = b

The general solution when \( a = b \) is: \( Y(x) = Ce^{-ax} + dx e^{-ax} \).

Analyze the Limit as x Approaches Infinity for a = b

As \( x \rightarrow \infty \), both terms \( Ce^{-ax} \) and \( dx e^{-ax} \) approach 0. Thus, \( \text{lim}_{x \rightarrow \infty} Y(x) = 0 \).

Key concepts

homogeneous equation

A homogeneous equation is a type of differential equation where all terms are dependent on the function and its derivatives. In our problem, the homogeneous part is: \(Y'(x) + aY(x) = 0\). This equation doesn't have any independent terms, meaning there's no constant or other function on the right side of the equation. To solve this equation, we can separate the variables and integrate. The solution is: \(Y_h(x) = Ce^{-ax}\), where \(C\) is an integration constant. This equation is critical because it helps us understand the behavior of the solutions as one part of the complete solution.

particular solution

A particular solution is a solution to the non-homogeneous differential equation that includes an additional function or term, not found in the homogeneous part. To find this, we assume a form that mimics the non-homogeneous term. In this case, for \(Y'(x) + aY(x) = de^{-bx}\), we guess a particular solution in the form \( Y_p(x) = A e^{-bx}\), substituting it into the equation. Simplifying, we discover that the coefficient \(A\) must satisfy \( A(a - b) = d \). So, \( A = \frac{d}{a-b} \), if \( a eq b \). Therefore, the particular solution is: \( Y_p(x) = \frac{d}{a-b} e^{-bx} \). This solution helps address the additional term in the original equation.

limit as x approaches infinity

To find the limit as \( x \) approaches infinity of a function, we need to analyze how each term in the solution behaves when \( x \) becomes large. For \( Y(x) = Ce^{-ax} + \frac{d}{a-b} e^{-bx} \), both exponential terms \( Ce^{-ax} \) and \( \frac{d}{a-b} e^{-bx} \) approach zero as \( x \) goes to infinity because both \( a \) and \( b \) are positive. Thus, \( \text{lim}_{x \rightarrow \text{infty}} Y(x) = 0 \). Even if \( a = b \), substituting into a different form, the separate terms \( Ce^{-ax} \) and \( dx e^{-ax} \) will still approach zero as \( x \) becomes very large because the exponential decay dominates the linear growth.

differential equation solution

A differential equation solution involves combining both the homogeneous and particular solutions to form the general solution. For our differential equation \(Y'(x) + aY(x) = de^{-bx}\), once we solve both parts individually, we add them together. If \( a eq b \), the solution is: \( Y(x) = Ce^{-ax} + \frac{d}{a-b} e^{-bx} \). If \( a = b \), after finding a particular solution \( dx e^{-ax} \), the general solution becomes: \( Y(x) = Ce^{-ax} + dx e^{-ax} \). This final solution represents all possible solutions of the differential equation based on different initial conditions. It encompasses both the behavior described by the homogeneous solution and modifications due to the non-homogeneous term.