Question

Solve \(y^{\prime}-y=e^{k t}\) with the initial condition \(y(0)=0 .\) As \(k\) approaches 1, what happens to your formula?

Short answer

As \( k \to 1 \), \( y(t) = t e^{t} \).

Step-by-step solution

Identify type of differential equation

The given differential equation is first-order linear and non-homogeneous: \( y' - y = e^{kt} \). This differential equation can be solved using the integrating factor technique.

Find the integrating factor

For the equation in the form \( y' + P(t)y = Q(t) \), the integrating factor \( \,\mu(t) \,\) is \( e^{\int P(t) \, dt} \). Rewrite \( y' - y = e^{kt} \) as \( y' + (-1)y = e^{kt} \), so \( P(t) = -1 \). The integrating factor is thus \( \mu(t) = e^{-t} \).

Solve the differential equation using the integrating factor

Multiply both sides of the equation \( y' - y = e^{kt} \) by the integrating factor \( e^{-t} \): \( e^{-t}y' - e^{-t}y = e^{kt}e^{-t} \). This simplifies to \( e^{-t}y' - e^{-t}y = e^{(k-1)t} \).Using the product rule, the left-hand side is the derivative of \( e^{-t}y \):\( \frac{d}{dt}(e^{-t}y) = e^{(k-1)t} \).Integrate both sides with respect to \( t \):\[ \int \frac{d}{dt}(e^{-t}y) dt = \int e^{(k-1)t} dt \]\( e^{-t}y = \frac{1}{k-1}e^{(k-1)t} + C \) where \( C \) is the constant of integration.

Solve for \( y(t) \)

Multiply through by \( e^{t} \) to solve for \( y(t) \):\( y = \frac{1}{k-1}e^{(k-1)t} e^{t} + Ce^{t} \).This simplifies further:\( y = \frac{1}{k-1}e^{kt} + Ce^{t} \).

Apply the initial condition

The initial condition is \( y(0) = 0 \). Substitute \( t = 0 \) into the equation:\( 0 = \frac{1}{k-1}e^{k \cdot 0} + C e^{0} \)\( 0 = \frac{1}{k-1} + C \)Thus, \( C = -\frac{1}{k-1} \).

Write the solution for \( y(t) \)

Substitute \( C = -\frac{1}{k-1} \) back into the equation for \( y(t) \):\( y(t) = \frac{1}{k-1}e^{kt} - \frac{1}{k-1}e^{t} \), which simplifies to:\( y(t) = \frac{1}{k-1}(e^{kt} - e^{t}) \).

Evaluate the limit as \( k \to 1 \)

As \( k \to 1 \), the expression \( \frac{1}{k-1}(e^{kt} - e^{t}) \) becomes undefined, so we use L'Hôpital's Rule:\[ \lim_{k \to 1} \frac{e^{kt} - e^{t}}{k-1} = \lim_{k \to 1} \frac{t e^{kt}}{1} = t e^{t} \]. So as \( k \) approaches 1, the solution becomes \( y(t) = t e^{t} \).

Key concepts

Integrating Factor

The integrating factor is a crucial concept when solving first-order linear differential equations. It serves as a multiplier that helps us transform a complicated differential equation into a simpler form. This simpler form is then more easily integrable, allowing us to solve the equation systematically.

In our given problem, the differential equation is expressed as \( y' - y = e^{kt} \). First, rewrite the equation in the standard form of first-order linear differential equation: \( y' + P(t)y = Q(t) \). Here, \( P(t) = -1 \).

To find the integrating factor, \( \mu(t) \), compute \( e^{\int P(t) \, dt} \). For \( P(t) = -1 \), the integrating factor becomes \( \mu(t) = e^{-t} \). By multiplying the entire differential equation by this integrating factor, the left side becomes the derivative of \( e^{-t}y \), simplifying the problem considerably.

This simplification allows us to integrate both sides more easily, leading to the solution for \( y \). Thus, the integrating factor plays a fundamental role in solving non-homogeneous differential equations like the one in our problem.

Non-homogeneous Differential Equation

A non-homogeneous differential equation includes a term that is not a function of the dependent variable alone; it also contains a non-zero external function that depends on the independent variable.First-order non-homogeneous differential equations generally have the form \( y' + P(t)y = Q(t) \), where \( Q(t) \) is the non-homogeneous term. In our exercise, \( e^{kt} \) is the non-homogeneous part.

These types of equations are distinct from homogeneous equations, which only have terms dependent on the unknown variable \( y \) and its derivatives.The presence of \( e^{kt} \) makes the direct integration of the equation impossible without manipulation through tools like the integrating factor.

Solving such a differential equation involves transforming it into a perfect derivative by using the integrating factor. We then integrate the resulting expression. The solution will include both the general solution of the homogeneous equation and a particular solution for the non-homogeneous part.

Initial Condition Problem

An initial condition problem provides additional information needed to determine the constant of integration when solving differential equations.This extra information is typically given in the form \( y(t_0) = y_0 \), where \( t_0 \) is a specific point and \( y_0 \) is the corresponding value of the solution. In this exercise, we have \( y(0) = 0 \).

After finding the general solution \( y(t) = C_1 e^{kt} + C_2 e^{t} \), we substitute the initial condition to find the constant \( C \). This process ensures that the specific solution satisfies the given initial condition.

It's vital to solve such problems accurately because initial conditions allow us to refine the broad general solutions into precise answers that conform to real-world scenarios or specific requirements in a mathematical context.In cases where \( k \) approaches a particular value, as seen in this problem when \( k = 1 \), special mathematical techniques such as L'Hôpital's Rule can be applied to resolve potentially undefined situations.