Question

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

Short answer

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

Step-by-step solution

Identify the unknown constant in the general solution

The general solution of the differential equation is given as \(y(t) = C e^{-3 t} + 10\). Here, the unknown constant is \(C.\)

Apply the initial condition

The initial condition is \(y(0) = 5\). To find the constant \(C\), we'll substitute \(t = 0\) into the general solution and then solve for \(C\): \(5 = C e^{-3(0)} + 10\)

Solve for the unknown constant

Now, we'll calculate the value of \(C\): \(5 = C e^{0} + 10\) \(5 = C(1) + 10\) \(C = 5 - 10\) \(C = -5\)

Write down the solution that satisfies the initial condition

With the computed value of \(C\), the solution that satisfies the initial condition \(y(0) = 5\) is: \(y(t) = -5 e^{-3 t} + 10\)

Key concepts

General Solution

A general solution to a differential equation includes all possible solutions. It is usually given in terms of an arbitrary constant, which is denoted by a symbol like \(C\). This constant allows the general solution to represent a family of curves or functions. The inclusion of the arbitrary constant is crucial because it reflects the nature of differential equations, which often have infinitely many solutions. Each value of \(C\) corresponds to a unique solution.In our exercise, the general solution is \(y(t) = C e^{-3t} + 10\). Here:

• \(e^{-3t}\) is a part of the general solution that handles the variable part based on the power of \(e\).
• The constant \(10\) is a particular solution that appears due to the setup of the equation or boundary conditions used to solve it.

To fully specify the solution, we must use additional information, like an initial condition, to determine \(C\). This completes the transition from a general to a specific solution.

Initial Condition

Initial conditions are essential for finding a specific solution from a general solution. They are conditions provided at a particular point, often at the start or at zero, which allow us to solve for the arbitrary constant in the general solution.In our example, the initial condition is \(y(0) = 5\). This means that when \(t = 0\), the function \(y(t)\) should equal 5. By substituting this into our general solution, \(y(t) = C e^{-3t} + 10\), and setting \(t = 0\), we get:\[y(0) = 5 = C e^{-3(0)} + 10\]Since \(e^{0} = 1\), this simplifies to:\[5 = C + 10\]Solving this equation gives \(C = -5\). Thus, the initial condition has enabled us to find the unique value of \(C\), leading to the particular solution \(y(t) = -5 e^{-3t} + 10\). This satisfies both the differential equation and the initial condition.

Exponential Function

The exponential function is one of the most important and widely used mathematical functions. It can be expressed as \(e^{x}\), where \(e\) (approximately 2.718) is the base of the natural logarithm. Exponential functions are characterized by their rapid growth or decay, depending on the sign of their exponent.In differential equations, exponential functions often appear in solutions because they naturally arise from processes with constant relative growth rates or decay rates.In the exercise, the term \(e^{-3t}\) is an exponential function with a decay rate. Here's why it's significant:

• The negative sign in the exponent \(-3t\) indicates exponential decay, meaning that as \(t\) increases, the value of \(e^{-3t}\) decreases towards zero.
• This function helps describe how the solution changes over time, based on the underlying differential equation dynamics.

Understanding exponential functions and their role in differential equations is key to solving and interpreting these mathematical problems. They describe how systems grow or shrink over time, offering insights into a wide range of real-world phenomena.