Problem 10 - Calculus: Early Transcendentals

Question

Solve the initial value problem. $$y^{\prime}+y \cos \left(e^{t}\right)=0, y(0)=0$$

Step-by-Step Solution

Answer

The solution is \( y(t) = 0 \) for all \( t \).

Step 1: Identify Equation Type

The given equation is a first-order linear differential equation of the form \( y' + P(t)y = 0 \), where \( P(t) = \cos(e^t) \).

Step 2: Solve the Homogeneous Equation

Since the equation is homogeneous, we can solve it by separation of variables. We rewrite the equation as \( y' = -y \cos(e^t) \).

Step 3: Separate Variables

Rewrite the equation as \( \frac{y'}{y} = -\cos(e^t) \). Integrate both sides: \( \int \frac{y'}{y} \, dt = \int -\cos(e^t) \, dt \).

Step 4: Integrate Both Sides

The integral of the left side is \( \ln |y| \), and for the right side, a substitution \( u = e^t \) gives \(-\int \cos(u) \, du = -\sin(u) + C'\). Therefore, \( \ln |y| = -\sin(e^t) + C'\).

Step 5: Solve for y

Exponentiate both sides to solve for \( y \): \( |y| = e^{-\sin(e^t) + C'} \). Let \( C = e^{C'} \), so \( y = Ce^{-\sin(e^t)} \).

Step 6: Apply Initial Condition

Given \( y(0) = 0 \). Substitute \( t = 0 \) into the equation: \( 0 = C e^{-\sin(e^0)} \). Since \( e^{-\sin(1)} eq 0 \), we must have \( C = 0 \).

Step 7: Determine Final Solution

With \( C = 0 \), the solution becomes \( y(t) = 0 \) for all \( t \).

Key Concepts

First-Order Linear Differential Equation

A first-order linear differential equation is an equation that involves a function and its first derivative. The standard form is:

• \( y' + P(t)y = Q(t) \)

In the given problem, the equation we are solving fits this definition. Here, \( P(t) = \cos(e^t) \) and \( Q(t) = 0 \).
The key characteristic is linearity, which means that the dependent variable \( y \) and its derivative \( y' \) are only to the first power (no exponents like \( y^2 \), or functions like \( \sin(y) \)).
Linear equations are significant because they have specific solution methods which guarantee a unique solution when initial conditions are provided.

Separation of Variables

Separation of variables is a method used to solve differential equations. It involves rearranging the equation such that each variable and its derivatives are on opposite sides of the equation.
This is particularly useful when solving homogeneous equations like our initial problem, where we could transform it into:

• \( \frac{y'}{y} = -\cos(e^t) \)

By separating variables, we can integrate both sides independently. The integrals will give us equations that relate \( y \) and \( t \), which we can solve to find the general solution. This method is quite powerful and often used in first-order differential equations.

Homogeneous Equation

A homogeneous equation is a type of differential equation where every term is a function of the dependent variable and its derivatives. For such equations, there is no standalone constant term, meaning \( Q(t) = 0 \).
For the given equation, the homogeneous form is:

• \( y' = -y \cos(e^t) \)

Homogeneous equations are simpler because the lack of a non-homogeneous part \( Q(t) \) means they can be solved directly using methods like separation of variables. These equations often represent processes that lose energy over time or decay without any external input.

Initial Condition

An initial condition specifies the value of the solution at a particular point, typically when \( t = 0 \). It helps in finding a specific solution from the general solution set.
In our problem, the initial condition was \( y(0) = 0 \).
This means that at time \( t = 0 \), the value of \( y \) must be zero. By substituting \( t = 0 \) into our solution, we use this condition to find the constant of integration. In this example, after applying the initial condition, we determined that \( C = 0 \).
This results in the particular solution: \( y(t) = 0 \) for all \( t \), which matches the initial condition perfectly.