Problem 4 - Calculus Early Transcendentals

Question

If the general solution of a differential equation is \(y=c e^{-3 t}+10,\) what is the solution that satisfies the initial condition \(y(0)=5 ?\)

Step-by-Step Solution

Answer

Answer: The particular solution of the given differential equation that satisfies the initial condition \(y(0) = 5\) is \(y = -5e^{-3t} + 10\).

Step 1: Identify the general solution

The given general solution of the differential equation is \(y = ce^{-3t} + 10\).

Step 2: Use the initial condition

We are given the initial condition \(y(0) = 5\). We can find the value of the constant 'c' by plugging in \(t=0\) and \(y(0)=5\) into the general solution: \(5 = c e^{-3(0)} + 10\)

Step 3: Solve for the constant 'c'

In the equation from Step 2, we can solve for 'c' as follows: \(5 = c e^{0} + 10 \Rightarrow 5 = c(1) + 10 \Rightarrow c = -5\)

Step 4: Write the particular solution

Now that we have found the value of 'c', we can substitute it back into the general solution to find the particular solution that satisfies the given initial condition: \(y = -5e^{-3t} + 10\) So, the solution of the differential equation that satisfies the initial condition \(y(0) = 5\) is \(y = -5e^{-3t} + 10\).

Key Concepts

General Solution

The general solution of a differential equation is the form that encapsulates all possible solutions for the equation. It typically includes one or more arbitrary constants, often denoted as 'c', which can take on any real value. This general solution arises because differential equations, much like indefinite integrals, have multiple solutions due to the presence of derivatives.

By including these constants, the general solution becomes a family of functions. Each function within this family is a potential solution to the differential equation. The presence of the constant accounts for the various shifts and transformations of the function, like scaling or translation, that still satisfy the equation.

In the exercise, the general solution is given as \( y = ce^{-3t} + 10 \). This expression shows that for any real number 'c', the equation will yield a solution that satisfies the differential equation. The constant 'c' is crucial as it determines the specific member of the family of possible solutions.

Initial Condition

An initial condition in the context of differential equations is a piece of information that allows us to pinpoint a unique solution from the general solution. It typically specifies the value of the function (and possibly its derivatives) at a particular point.

Using initial conditions is necessary to transition from the general solution, which contains arbitrary constants, to a specific solution that fits the exact scenario described by the problem.

Consider the initial condition in our exercise: \( y(0) = 5 \). This tells us that when \( t = 0 \), the function \( y \) must equal 5. By substituting \( t = 0 \) into the general solution \( y = ce^{-3t} + 10 \), we solve for the constant 'c'. This step is pivotal in determining the exact curve from the family of curves that the general solution represents, meeting the initial criteria given by the problem.

Particular Solution

A particular solution to a differential equation is a specific function that satisfies both the differential equation and the initial conditions provided. Once the constant(s) in the general solution are determined using the initial condition(s), substituting them back gives us the particular solution.

The process of finding a particular solution boils down to eliminating the arbitrary constants in the general solution, making it unique and tailor-made for the problem at hand.

In the given exercise, after finding the value of \( c = -5 \) using the initial condition \( y(0) = 5 \), we substitute it into our general solution to get the particular solution: \( y = -5e^{-3t} + 10 \).

This expression is the specific solution that fits both the differential equation and the conditions defined by the particular scenario, thereby providing the precise answer required by the task.