Question

The general solution of a first-order linear differential equation is \(y(t)=C e^{-10 t}-13 .\) What solution satisfies the initial condition \(\bar{y}(0)=4 ?\)

Short answer

Question: Determine the particular solution to the first-order linear differential equation y(t) = Ce^{-10t} - 13 with the initial condition y(0)=4. Solution: The particular solution that satisfies the initial condition is y(t) = 17e^{-10t} - 13.

Step-by-step solution

Write down the general solution and initial condition

The general solution of the given first-order linear differential equation is \(y(t) = Ce^{-10t} - 13\). The initial condition is \(\bar{y}(0) = 4\).

Substitute the initial condition into the general solution

To find the particular solution satisfying the initial condition, let's substitute \(\bar{y}(0) = 4\) in the given general solution: $$4 = Ce^{-10(0)} - 13.$$

Solve for the constant C

Now, we will solve the equation from Step 2 for the constant C. Since \(e^{-10(0)} = e^0 = 1\), the equation becomes: $$4 = C(1) - 13.$$ Adding 13 to both sides, we get $$C = 4 + 13 = 17.$$

Find the particular solution

Now that we have found the constant C, we can plug its value back into the general solution to find the particular solution that satisfies the initial condition: $$y(t) = 17e^{-10t} - 13.$$

Key concepts

First-Order Differential Equation

A first-order differential equation is an equation that involves the first derivative of an unknown function and the function itself. These equations are foundational in mathematics because they can describe a wide range of physical phenomena, from population growth to electrical circuits. In our exercise, the given first-order differential equation is expressed in its general solution form:
\[ y(t) = C e^{-10t} - 13 \]
This equation involves the constant \(C\), which is an arbitrary constant, and the function \(e^{-10t}\), which represents an exponential decay.

The components of the equation include:

• The exponential term \(e^{-10t}\) which indicates the rate at which the function decreases as \(t\) increases.
• The constant \(-13\) acts like a shift, which moves the entire graph of the function downward by 13 units on the y-axis.

First-order equations are particularly interesting because they're simple yet powerful. Solving them typically involves finding the value of the constant to tailor a solution to meet specific conditions.

Initial Conditions

Initial conditions are critical pieces of information that help us determine a specific solution from a set of possible solutions provided by a differential equation. They're like a starting point that discerns which path the function should follow.

In the context of our example, we're given the initial condition \( \bar{y}(0) = 4\). This tells us that when the function is evaluated at \( t = 0 \), its output should be 4.
Applying this in our general solution, we substitute \( t = 0 \) into the equation:
\[ 4 = Ce^{-10(0)} - 13. \]
Since \( e^{0} = 1 \), the equation simplifies and allows us to solve for the unknown constant \(C\).

Initial conditions play an important role in not only pinpointing a unique solution but also in modeling real-world situations where initial values are known, such as the initial velocity of a traveling object.

Particular Solution

The particular solution of a differential equation is the specific solution that satisfies both the differential equation and any given initial conditions. Once we have found \(C\) through the initial conditions, we can replace it back into the general solution.

In our example, after solving for \(C\) using the initial condition, we found \(C = 17\). Substituting \(C = 17\) in the general solution gives us:
\[ y(t) = 17e^{-10t} - 13. \]
This equation is now tailored to meet the initial condition \( \bar{y}(0) = 4 \) precisely, and it represents a function that starts at y = 4 when t = 0, evolving over time.

The particular solution is unique and provides exact information tailored to scenarios where exact initial values are important. It's a way of reconciling the theoretical model (the general solution) with particular real-world data (the initial conditions). This interaction makes differential equations a powerful tool in both mathematics and applied sciences.