Problem 16

Question

Solve the given autonomous differential equations. $\frac{d N}{d t}=5-N$, where $N(2)=3$

Step-by-Step Solution

Verified

Answer

The solution is $ N(t) = -2e^{2-t} + 5 $.

1Step 1: Identify the Differential Equation Type

The given differential equation is $ \frac{dN}{dt} = 5 - N $. This is a first-order linear ordinary differential equation with constant coefficients.

2Step 2: Solve the Homogeneous Equation

First, solve the homogeneous part: $ \frac{dN}{dt} = -N $. This can be rewritten as $ \frac{dN}{N} = -dt $. Integrating both sides gives $ \ln|N| = -t + C_1 $, or solving for $ N $: $ N = C_1 e^{-t} $.

3Step 3: Find a Particular Solution

For the non-homogeneous part, assume a particular solution $ N_p = C $. Plugging $ N_p = C $ into the differential equation $ \frac{dN}{dt} = 5 - N $ gives $ 0 = 5 - C $. Thus, $ C = 5 $, so $ N_p = 5 $.

4Step 4: Write the General Solution

The general solution is the sum of the homogeneous solution and the particular solution: $ N(t) = C_1 e^{-t} + 5 $.

5Step 5: Use Initial Condition to Find Constant

The initial condition is $ N(2) = 3 $. Substitute into the general solution: $ 3 = C_1 e^{-2} + 5 $. Solving for $ C_1 $, we get $ C_1 e^{-2} = -2 $, hence $ C_1 = -2e^2 $.

6Step 6: Final Solution

Substitute $ C_1 $ back into the general solution: $ N(t) = -2e^{2-t} + 5 $.

Key Concepts

First-order Linear Ordinary Differential EquationHomogeneous SolutionParticular SolutionInitial Value Problem

First-order Linear Ordinary Differential Equation

A first-order linear ordinary differential equation is a type of equation that involves derivatives of a function with respect to one variable, usually time or space. Specifically, it has the form:

$\frac{dy}{dt} + p(t)y = g(t)$

This equation is characterized by being "linear," which means the unknown function and its derivative are not raised to any power other than 1. Moreover, "first-order" indicates that the highest derivative is the first derivative.
In the case of the equation $\frac{dN}{dt} = 5 - N$, it's clear that this is a first-order linear differential equation since it involves the first derivative $\frac{dN}{dt}$, and both $N$ and its derivative are of the first power. It's also autonomous, meaning it doesn't explicitly depend on the independent variable $t$. This allows us to solve it using techniques like separation of variables and integrating.

Homogeneous Solution

The homogeneous solution is an important part of the strategy to solve differential equations. When dealing with a first-order linear equation, finding the homogeneous solution involves setting the equation to zero:

$\frac{dN}{dt} = -N$

This is the homogeneous equation corresponding to our original problem. To solve this, we can use separation of variables by moving all terms involving $N$ to one side:

$\frac{dN}{N} = -dt$

Integrating both sides, we obtain:

$\ln|N| = -t + C_1$

Solving for $N$, we find the homogeneous solution:

$N = C_1 e^{-t}$

This expression describes the natural response of the system without external influences.

Particular Solution

Apart from the homogeneous solution, solving a first-order linear ordinary differential equation also involves finding a particular solution. This accounts for the non-homogeneous part of the equation. In our problem, we assume a particular solution of the form:

$N_p = C$

Substituting this constant into the original differential equation:

$\frac{dN}{dt} = 5 - C$

Since the derivative of a constant is zero, we get:

$0 = 5 - C$

This equation shows that $C = 5$. Thus, $N_p = 5$ is a particular solution, indicating a steady state where the population remains constant.

Initial Value Problem

An initial value problem (IVP) specifies not only the differential equation but also an initial condition that the solution must satisfy. This condition lets us find the specific solution of the differential equation that fits the situation. In our example, the initial condition given is: