Question

Substitute \(y=B e^{3 t}\) into \(y^{\prime}-y=8 e^{3 t}\) to find a particular solution.

Short answer

The particular solution is \(y = 4 e^{3t}\).

Step-by-step solution

Write down the given substitution

The problem gives a substitution for the function \(y\) in the differential equation: \(y = B e^{3t}\). Substitute this into the equation \(y' - y = 8e^{3t}\).

Differentiate the substituted function

Differentiate \(y = B e^{3t}\) with respect to \(t\). The derivative \(y'\) is given by \(y' = \frac{d}{dt}(B e^{3t}) = 3B e^{3t}\).

Substitute back into the original differential equation

Replace \(y'\) and \(y\) in the equation \(y' - y = 8e^{3t}\) with the expressions obtained from substitution and differentiation: \(3B e^{3t} - B e^{3t} = 8 e^{3t}\).

Simplify the equation

Simplify the left-hand side: This becomes \(2B e^{3t}\) because \(3B e^{3t} - B e^{3t} = 2B e^{3t}\). So, the equation becomes \(2B e^{3t} = 8 e^{3t}\).

Solve for B

Since the terms involving \(e^{3t}\) are on both sides, you can divide by \(e^{3t}\), giving \(2B = 8\). Therefore, solve for \(B\) to get \(B = 4\).

Write the particular solution

Substitute the value found for \(B\) back into \(y = B e^{3t}\) to find the particular solution: \(y = 4 e^{3t}\).

Key concepts

Particular Solution

In differential equations, finding a particular solution is about discovering a specific solution that satisfies both the differential equation and any given initial conditions. This is opposed to the general solution, which encompasses all possible solutions to the differential equation.

By applying a particular solution, we pinpoint the unique function that aligns with additional constraints.

• The process often involves guessing or using a provided form, like in this exercise where we're given the substitution function, such as \(y = B e^{3t}\).
• The goal is to find the constant \(B\) that satisfies the given relation, here, \(y' - y = 8 e^{3t}\).

Using this substitute function, we can integrate any specific conditions of a problem to solve for unknowns like \(B\), which gives us meaningful results within its context.

Differentiation

Differentiation is a fundamental concept in calculus and a key tool in working with differential equations. When differentiating, you're finding the derivative of a function, which is a way of determining its rate of change.

In the context of our exercise, differentiation is used to handle the given function \(y = B e^{3t}\). Let's delve into its steps:

• Differentiate using the chain rule: The derivative of \(e^{3t}\) with respect to \(t\) is \(3e^{3t}\) because of the exponent's multiplier.
• As such, differentiating \(y\) gives \(y' = \frac{d}{dt}(B e^{3t}) = 3B e^{3t}\).

This process is crucial as it prepares the function to be used within the larger differential equation, allowing for substitution that aligns with the differential relationship presented.

Exponential Functions

Exponential functions are mathematical functions of the form \(f(t) = a e^{kt}\), where base \(e\) is a constant approximately equal to 2.718, and \(a\) and \(k\) are constants. They play a significant role in growth and decay processes that occur at a constant rate relative to their size.

In the exercise, the function \(y = B e^{3t}\) is exponential. Here's how they integrate into differential equations:

• They have a unique property where their derivatives are proportional to the function itself, making them "natural" solutions to many differential equations.
• In \(y = B e^{3t}\), when differentiated, it maintains its exponential form, which greatly simplifies solving the differential equations.

Due to these attributes, exponential functions become very useful in modeling natural growth and decay scenarios, such as population dynamics and radioactive decay, among others.