Question

Show that if \(a\) and \(\lambda\) are positive constants, and \(b\) is any real number, then every solution of the equation $$ y^{\prime}+a y=b e^{-\lambda t} $$ has the property that \(y \rightarrow 0\) as \(t \rightarrow \infty\) Hint: Consider the cases \(a=\lambda\) and \(a \neq \lambda\) separately.

Short answer

Question: For the given first-order linear ordinary differential equation \(y' + ay = be^{-\lambda t}\), show that every solution has the property that \(y \rightarrow 0\) as \(t \rightarrow \infty\). Answer: In both cases (\(a=\lambda\) and \(a\neq\lambda\)), the solutions of the given ODE have the desired property, as shown in the step-by-step solution. Specifically, for \(a = \lambda\), the solution is \(y(t) = \frac{bt + C}{e^{at}}\), and for \(a \neq \lambda\), the solution is \(y(t) = \frac{b}{a-\lambda}e^{-(\lambda - a)t} + De^{-at}\). As \(t \rightarrow \infty\), both expressions approach 0, fulfilling the condition \(y \rightarrow 0\).

Step-by-step solution

Case \(a=\lambda\)

When \(a = \lambda\), our equation becomes: $$ y' + ay = be^{-at} $$ Now, let's solve this first-order linear ODE using an integrating factor. The integrating factor is given by \(I(t) = e^{\int a dt} = e^{at}\).

Multiply the equation by the integrating factor

Multiply both sides of the equation by the integrating factor, \(e^{at}\): $$ e^{at}(y' + ay) = be^{at}e^{-at} $$ The left-hand side of the equation now becomes the derivative of the product of \(y(t)\) and the integrating factor, \(e^{at}\): $$ \frac{d}{dt}(e^{at}y) = b $$

Integrate both sides of the equation

Now, let's integrate both sides of the equation with respect to \(t\): $$ \int \frac{d}{dt}(e^{at}y) dt = \int b dt $$ On the left-hand side, we can cancel the derivative with the integral, which leaves us with \(e^{at}y\). On the right-hand side, the integral of a constant \(b\) with respect to \(t\) is \(bt + C\), where \(C\) is the integration constant: $$ e^{at}y = bt + C $$

Solve for \(y(t)\)

Now, let's solve for \(y(t)\) by dividing both sides by \(e^{at}\): $$ y(t) = \frac{bt + C}{e^{at}} $$ As \(t \rightarrow \infty\), we can see that the denominator \(e^{at}\) grows without bound, so \(y(t) \rightarrow 0\).

Case \(a\neq\lambda\)

When \(a \neq \lambda\), our equation remains as it was given: $$ y' + ay = be^{-\lambda t} $$ Again, let's solve this first-order linear ODE using an integrating factor. The integrating factor is still given by \(I(t) = e^{\int a dt} = e^{at}\).

Multiply the equation by the integrating factor

Multiply both sides of the equation by the integrating factor, \(e^{at}\): $$ e^{at}(y' + ay) = be^{at}e^{-\lambda t} $$ The left-hand side of the equation now becomes the derivative of the product of \(y(t)\) and the integrating factor, \(e^{at}\): $$ \frac{d}{dt}(e^{at}y) = be^{(a-\lambda)t} $$

Integrate both sides of the equation

Next, let's integrate both sides of the equation with respect to \(t\): $$ \int \frac{d}{dt}(e^{at}y) dt = \int be^{(a-\lambda)t} dt $$ On the left-hand side, we can cancel the derivative with the integral, which leaves us with \(e^{at}y\). On the right-hand side, we have: $$ \int be^{(a-\lambda)t} dt = \frac{b}{a-\lambda}e^{(a-\lambda)t} + D $$ where \(D\) is another integration constant.

Solve for \(y(t)\)

Now, let's solve for \(y(t)\) by dividing both sides by \(e^{at}\): $$ y(t) = \frac{b}{a-\lambda}e^{(a-\lambda)t}e^{-at} + De^{-at} $$ Rewriting the equation gives: $$ y(t) = \frac{b}{a-\lambda}e^{-(\lambda - a)t} + De^{-at} $$ As \(t \rightarrow \infty\), the term \(e^{-(\lambda - a)t}\) approaches 0, since \(\lambda \neq a\). Similarly, the term \(e^{-at}\) becomes 0 as well. Therefore, we have \(y(t) \rightarrow 0\) in this case as well. In conclusion, for both cases (\(a=\lambda\) and \(a\neq\lambda\)), the solutions of the given ODE have the property \(y \rightarrow 0\) as \(t \rightarrow \infty\).

Key concepts

First-Order Linear ODE

A first-order linear ordinary differential equation (ODE) is a type of differential equation that involves a first derivative of a function. In the form \( y' + P(t) y = Q(t) \), it is linear with respect to \( y \). Such equations are common in various scientific fields, providing a tool to model relationships where changes over time are linear. In the exercise, the equation \( y^{\prime}+a y = b e^{-\lambda t} \) fits this pattern.
To solve first-order linear ODEs, we typically use integrating factors or substitution methods. The importance lies in their ability to approximate real-world phenomena such as electrical circuits, population growth, or decay processes. The linearity implies proportionality and simplicity but may not capture all complex dynamics, being effective within certain bounds of initial conditions and linear behaviors.

Integrating Factor

An integrating factor is a function used to solve linear differential equations more easily. It multiplies every term in the equation, simplifying it into a format that integrates cleanly. For a given differential equation \( y' + P(t) y = Q(t) \), the integrating factor is given by \( \mu(t) = e^{\int P(t) dt} \).

• The goal is to transform the left-hand side of the equation into a derivative of a product.
• In the exercise provided, the integrating factor \( I(t) = e^{at} \) allows for the equation to be rewritten as the derivative \( \frac{d}{dt}(e^{at}y) \).

By solving the resulting equation, integration becomes straightforward, leading to expressions where continuous and differentiable functions can be solved in terms of elementary functions, often resulting in straightforward expressions for the solution \( y(t) \). The clarity of this process highlights why integrating factors are preferred in solving linear ODEs.

Limit as \(t \rightarrow \infty\)

Understanding the behavior of solutions as time \( t \) moves towards infinity helps us foresee long-term trends in models and equations. The asymptotic behavior analyzed, as shown in the step-by-step solution, determines the ultimate disposition of functions in time.

• In our exercise, regardless of whether \( a = \lambda \) or \( a eq \lambda \), the solution \( y(t) \) approaches \( 0 \) as \( t \rightarrow \infty \).
• It is a crucial property for confirming if a model naturally decomposes over time or if it will stabilize to a non-zero state.

Such analyses are vital in areas like physics (for understanding damped motion), biological systems (for determining long-term rates of reaction), and economics (for evaluating the effect of decay rates on investments). Mathematical rigor in proving these limits helps ensure models' reliability and applicability in forecasting scenarios.

Exponential Decay

Exponential decay describes processes where the quantity reduces at a rate proportional to its current value. It is one of the key results in analyzing differential equations where coefficients remain constant.

• In our exercise, when computing \( y(t) \, ext{as} \, t \rightarrow \infty \), terms like \( e^{at} \) and \( e^{-(\lambda - a)t} \) emerge.
• These terms decay exponentially if the exponent is negative, steadily driving \( y(t) \) towards zero over time.

The phenomenon is frequently encountered in physics, such as radioactive decay or thermal cooling and is adjustable to encompass more complex processes, modifying assumptions to fit various decay or growth rates. It highlights the practical, intuitive understandings in mathematics, emphasizing how initial parameters critically determine time evolution and eventual equilibrium states inside controlled systems.