Problem 30 - Calculus

Question

(a) Solve the differential equation $$\frac{d P}{d t}=0.2 P-10$$ Write the solution \(P\) as an explicit function of \(t\) (b) Find the particular solution for each initial condition below and graph the three solutions on the same coordinate plane. $$P(0)=40, \quad P(0)=50, \quad P(0)=60$$

Step-by-Step Solution

Answer

The general solution is \(P = 5Ce^{0.2t} + 50\); particular solutions are \(P(t) = -10e^{0.2t} + 50\), \(P(t) = 50\), and \(P(t) = 10e^{0.2t} + 50\).

Step 1: Separate Variables

The given differential equation is \( \frac{dP}{dt} = 0.2P - 10 \). To solve it, we first separate the variables. Rewriting, we get:\[\frac{dP}{0.2P - 10} = dt\]

Step 2: Integrate Both Sides

Integrate both sides of the equation. On the left side, integrate with respect to \(P\), and on the right side, integrate with respect to \(t\):\[\int \frac{dP}{0.2P - 10} = \int dt\]The left side is a natural logarithm integral, and the right side is a simple integration:\[\frac{1}{0.2} \ln|0.2P - 10| = t + C\]

Step 3: Solve for ln Expression

Multiply through by 0.2 to eliminate the fraction:\[\ln|0.2P - 10| = 0.2t + C\]

Step 4: Exponentiate to Solve for P

Exponentiate both sides to solve for \(P\):\[|0.2P - 10| = e^{0.2t + C}\]This simplifies to:\[0.2P - 10 = Ce^{0.2t}\]where \(C\) is a new constant obtained from \(e^C\).

Step 5: Solve for P

Rearrange to solve for \(P\):\[0.2P = Ce^{0.2t} + 10\]\[P = 5Ce^{0.2t} + 50\]

Step 6: Determine Particular Solutions

Substitute the initial conditions to find specific solutions. For \(P(0) = 40\):\[ 40 = 5C + 50 \Rightarrow C = -2 \]So, \( P(t) = -10e^{0.2t} + 50 \).For \(P(0) = 50\):\[ 50 = 5C + 50 \Rightarrow C = 0 \]So, \( P(t) = 50 \).For \(P(0) = 60\):\[ 60 = 5C + 50 \Rightarrow C = 2 \]So, \( P(t) = 10e^{0.2t} + 50 \).

Step 7: Graph the Solutions

Graph the three solutions:- \(P(t) = -10e^{0.2t} + 50\)- \(P(t) = 50\)- \(P(t) = 10e^{0.2t} + 50\)Each solution represents different initial values presented as separate curves on the coordinate plane.

Key Concepts

Separation of Variables

Separation of Variables is a common technique used to solve first-order differential equations. The goal of this method is to rearrange the equation so that each side contains only one variable. This allows us to integrate both sides separately.
Let's take the differential equation from our exercise: \( \frac{dP}{dt} = 0.2P - 10 \). To separate the variables, we move all terms involving \(P\) to one side and \(dt\) to the other. This results in:

• \( \frac{dP}{0.2P - 10} = dt \)

Now, both sides of the equation can be integrated individually. This technique is valuable because it simplifies the process of solving differential equations step by step.
The integral of the left side depends exclusively on \(P\), while the right side depends exclusively on \(t\). Ultimately, separation of variables makes it easier to find the general solution of the differential equation.

Initial Conditions

Initial conditions are specific values given at a certain point, often at \(t=0\), to uniquely determine the particular solution of a differential equation. They ensure that out of all possible solutions, the one that satisfies the initial condition is chosen.
In our exercise, we have three initial conditions:

• \( P(0) = 40 \)
• \( P(0) = 50 \)
• \( P(0) = 60 \)

These initial conditions are crucial for calculating the constant \(C\) in the general solution. By substituting these values into the general form of the solution, we can find specific solutions for \(P(t)\). Each initial condition gives us a unique solution that passes through the specified point on the \(t-P\) plane.
This helps in modeling specific scenarios or initial states of the system described by the differential equation.

Exponential Function

The exponential function often appears when solving differential equations, especially in cases involving growth or decay. This function is characterized by the base \(e\), where the equation resembles \(e^{kt}\) with \(k\) as a constant.
In our solution:

• \[ P = 5Ce^{0.2t} + 50 \]

Here, \(e^{0.2t}\) is the exponential term. This term represents the dynamic change over time according to the rate \(0.2\). The presence of \(e\) indicates continuous compounding over time, reflecting realistic scenarios such as population growth or decline.
Exponential functions are fundamental in differential equations due to their unique properties of differentiation and integration, which tend to yield the function itself or a simple scaled version of it. This self-replicating nature eases the progression from differential equations to solutions capturing an underlying process.

Particular Solution

A particular solution is a specific solution to a differential equation that satisfies given initial conditions. It differs from the general solution, which includes unspecified constants.
In the context of our problem, starting from the general solution:

• \[ P(t) = 5Ce^{0.2t} + 50 \]

Applying initial conditions such as \( P(0) = 40 \), we find \( C = -2 \) resulting in the particular solution \( P(t) = -10e^{0.2t} + 50 \). This gives us a curve that fits the initial state precisely, ensuring accurate modeling of the scenario.
Each different initial condition may lead to a distinct particular solution. The process involves substituting these conditions into the general solution to isolate the constant, ultimately pinning down the unique solution that fits specific criteria.