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In Problems , solve each autonomous differential equc tion. $$ \frac{d N}{d t}=5-N, \text { where } N(2)=2 $$

Short Answer

Expert verified
The solution is \( N(t) = 5 - 3e^{2-t} \).

Step by step solution

01

Separate Variables

The given autonomous differential equation is \( \frac{dN}{dt} = 5 - N \). This equation can be rewritten in a form that allows separation of variables: \( \frac{dN}{5-N} = dt \). By separating the variables, we aim to get everything involving \( N \) on one side and everything involving \( t \) on the other.
02

Integrate Both Sides

Now, integrate both sides of the equation: \( \int \frac{1}{5-N} \, dN = \int 1 \, dt \). The left side integrates to yield \(-\ln|5-N|\), and the right side integrates to \(t + C\), where \(C\) is the constant of integration. Thus, we have: \(-\ln|5-N| = t + C \).
03

Solve for the Integration Constant

We solve for \( C \) using the initial condition \( N(2) = 2 \). Substitute \( t = 2 \) and \( N = 2 \) into the equation: \(-\ln|5-2| = 2 + C\). Simplifying yields \(-\ln 3 = 2 + C\), so \( C = -\ln 3 - 2 \).
04

Solve for N

Now substitute \( C \) back into the equation: \(-\ln|5-N| = t - \ln 3 - 2 \). Multiply by -1 to get \( \ln|5-N| = -t + \ln 3 + 2 \). Exponentiate both sides to eliminate the natural logarithm: \(|5-N| = e^{-t + \ln 3 + 2} \). Rewrite using property \( e^{a+b} = e^a e^b \): \( |5-N| = 3e^{2}e^{-t} \).
05

Consider the Absolute Value

Since \( N \) must reflect the initial condition and the function \( 5 - N \), \( N \) is less than 5 initially. Thus, resolve the absolute value: \( 5-N = 3e^{2}e^{-t} \) leads to \( N = 5 - 3e^{2}e^{-t} \).
06

Simplify and Find the Particular Solution

The particular solution that satisfies the initial condition \( N(2) = 2 \) is now simply \( N = 5 - 3e^{2-t} \), confirming that substitution of \( t=2 \) gives \( N=2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Separation of Variables
The technique of separating variables is a fundamental method used to solve differential equations. Especially useful in solving autonomous differential equations, this method works by rearranging the equation so that each variable appears on a separate side of the equation. In our problem with the differential equation \(\frac{dN}{dt} = 5 - N\), separation of variables involves rewriting it as \(\frac{dN}{5-N} = dt\). This structure allows us to integrate both sides independently, which is crucial for solving the equation step by step.

By separating the variables, we isolate \(N\)-related terms from \(t\)-related ones. This process prepares the equation for integration. Once separated, the differential equation becomes easier to handle, making it possible to find a solution by subsequent integration.
Initial Condition
The initial condition in a differential equation provides essential information needed to find a particular solution. It defines a specific point in time (or space) where the value of the solution is known. Initial conditions are critically important because many differential equations have an infinite number of solutions; the initial condition helps pin down which one is relevant to the given situation.

In our example, the initial condition is given as \(N(2) = 2\). This means that when \(t = 2\), the value of \(N\) is precisely 2. By using this information, we can determine the integration constant after performing the integration step. By substituting \(t=2\) and \(N=2\) into the integrated equation, we solve for the constant that makes the equation conform to this specific solution behavior.
Particular Solution
A particular solution of a differential equation is a solution that satisfies both the differential equation and the initial conditions. It is not just any solution; it is the specific one that fits within the unique constraints of the problem.

For our autonomous differential equation, after performing integration and incorporating the initial condition, we derive the particular solution. In this case, the solution \(N = 5 - 3e^{2-t}\) fits both the differential relationship and the initial condition. This solution perfectly describes how the quantity \(N\) evolves over time \(t\) in alignment with the initial condition when \(t=2, N=2\).
  • It provides not just any trajectory but the precise path \(N\) follows under the given circumstances.
  • This particular solution is vital because it ensures that the modeled behavior accurately reflects the real-world system described by the initial condition.
Integration Constant
The integration constant, represented often by \(C\), emerges naturally from the process of integrating a differential equation. Since indefinite integration results in a family of curves or functions differing by a constant, we introduce \(C\) to account for this variability.

In solving \(\int \frac{1}{5-N} \, dN = \int 1 \, dt\), following integration, an expression involving \(C\) appears: \(-\ln|5-N| = t + C\). The role of \(C\) is crucial because it allows us to modify the generic solution such that it passes through the specific initial condition provided.

Using the initial condition \(N(2) = 2\), we calculated \(C\) as \(-\ln 3 - 2\). This calculation is a key step in personalizing the solution, ensuring it is not just a generic solution but one tailored exactly to fit the given conditions defined by the problem.

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Most popular questions from this chapter

In our compartment model we assumed that inflows and outflows are matched at \(q\) to keep the volume of water in the tank constant. It's often useful when modeling, for example, the flow of pollutant into a pristine environment, to consider what can occur if the inflows and outflows do not match. Let's assume that the tank initially contains a volume \(V_{0}\) of water. Water flows into the tank at rate \(q_{\mathrm{in}}\), and out of the tank at rate \(q_{\text {out. }}\) (You may assume \(q_{\text {in }}>q_{\text {out } .}\) ) Suppose that the water flowing into the tank contains a concentration \(C_{I}\) of solute. As usual we write \(C(t)\) for the concentration in the tank. (a) Show that the concentration in the tank can be modeled using a differential equation: $$ \frac{d}{d t}(C V)=q_{\text {in }} C_{I}-q_{\text {out }} C $$ (b) Previously we were able to treat \(V\) as a constant. Now \(V\) changes with time. Derive a formula for \(V(t)\). (c) By substituting your formula for \(V(t)\) into (a), derive a differential equation for \(C(t)\). (d) In general we cannot analyze the behavior of the solution \(C(t)\) using techniques from Section \(8.2 .\) Why not? (e) Let's assume \(C_{l}=0\). Then show that your equation from (c) can be written as: $$ \frac{d C}{d t}=\frac{-q_{\mathrm{in}} C}{V_{0}+\left(q_{\mathrm{in}}-q_{\mathrm{out}}\right) t} $$ (f) Assume some definite values for the constants in \((8.57):\) \(q_{\mathrm{in}}=2, q_{\text {out }}=1\), and \(V_{0}=20 .\) Assuming \(C(0)=1\), solve \((8.57)\) to find \(C(t) .\) Show that \(\lim _{t \rightarrow \infty} C(t)=0\).

Determine whether the equilibrium at \(x=0\) is stable, unstable, or semi- stable. $$ \frac{d x}{d t}=x^{2}-x^{3} $$

Find all equilibria, and, by calculating the eigenvalue of the differential equation, determine which equilibria are stable and which are unstable. $$ \frac{d y}{d x}=\frac{1}{y^{3}}-\frac{1}{y} \quad, \quad y>0 $$

Find all equilibria, and, by calculating the eigenvalue of the differential equation, determine which equilibria are stable and which are unstable. \(\frac{d x}{d t}=h x-x^{2}\), where \(h\) is a constant and (a) \(h>0\), (b) \(h<0\)

Find the equilibria of the following differential equations. $$ \frac{d y}{d t}=\frac{y-1}{y^{2}+1} $$

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