Problem 4 - Applied Calculus

Question

Find particular solutions. $$\frac{d m}{d t}=0.1 m+200, \quad m(0)=1000$$

Step-by-Step Solution

Answer

The particular solution is \( m(t) = -2000 + 3000e^{0.1t} \).

Step 1: Identify the Type of Differential Equation

Firstly, notice that the differential equation \( \frac{d m}{d t} = 0.1m + 200 \) is a first-order linear differential equation. This is because it contains the first derivative \( \frac{d m}{d t} \), plus a term that affects \( m \) linearly.

Step 2: Write the Equation in Standard Form

The standard form of a linear differential equation is \( \frac{d m}{d t} - P(t)m = Q(t) \). Rearrange the given equation to this form:\[ \frac{d m}{d t} - 0.1m = 200 \]

Step 3: Find the Integrating Factor

The integrating factor \( \mu(t) \) is given by \( e^{\int -0.1 \, dt} \). Calculate this integral:\[ \int -0.1 \, dt = -0.1t \]Thus, the integrating factor is:\[ \mu(t) = e^{-0.1t} \]

Step 4: Multiply through by the Integrating Factor

Multiply every term in the equation by the integrating factor:\[ e^{-0.1t} \cdot \frac{d m}{d t} - e^{-0.1t} \cdot 0.1m = 200 \cdot e^{-0.1t} \]This transforms the left side into the derivative of the product \( \frac{d}{dt}(e^{-0.1t}m) \).

Step 5: Integrate Both Sides

Integrate both sides with respect to \( t \):\[ \int \frac{d}{dt}(e^{-0.1t}m) \, dt = \int 200 e^{-0.1t} \, dt \]This results in:\[ e^{-0.1t}m = -2000 e^{-0.1t} + C \]

Step 6: Solve for \( m(t) \)

Solve the equation for \( m \):\[ m(t) = -2000 + Ce^{0.1t} \]

Step 7: Apply Initial Condition

Use the initial condition \( m(0) = 1000 \) to solve for the constant \( C \):\[ 1000 = -2000 + C e^{0} \]\[ C = 3000 \]

Step 8: Write the Particular Solution

Substitute \( C = 3000 \) back into the expression for \( m(t) \):\[ m(t) = -2000 + 3000e^{0.1t} \]

Key Concepts

Integrating Factor

When solving first-order linear differential equations, the concept of the integrating factor becomes indispensable. It serves as a tool to simplify the equation and make it solvable.

To find the integrating factor, denoted often as \( \mu(t) \), you have to look at the standard form of the equation: \( \frac{d m}{d t} - P(t)m = Q(t) \). The expression \( P(t) \) dictates the integrating factor:

• Compute the integral of \(-P(t)\)
• Exponentiate the result to find \( \mu(t) = e^{\int -P(t) \, dt} \)

In the given example, \( P(t) = 0.1 \). The integrating factor then becomes \( e^{-0.1t} \).

Once you find this factor, multiply the entire original equation by it. This step conveniently transforms the equation into a form where the left side becomes the derivative of a product of functions, making integration straightforward.

Initial Conditions

Initial conditions are crucial in determining a unique solution for a differential equation. They provide the necessary extra information to pinpoint one specific solution out of many possibilities.

In differential equations, if you're given a condition such as \( m(0) = 1000 \), it means that when the variable, often time \( t \), is zero, the function \( m(t) \) equals 1000. Such initial conditions help determine any unknown constants that appear in your general solution.

• Substitute the initial condition into your general solution.
• Solve the equation for the unknown constant \( C \).

For instance, after integrating in the example exercise, the initial condition \( m(0) = 1000 \) was used to find \( C = 3000 \). This step transforms a general solution into a particular one.

Particular Solution

Finding the particular solution to a differential equation is about homing in on one precise answer satisfying given initial conditions. After guiding through the steps of solving, merging the general form with initial conditions gives a particular form.

Once you've solved the differential equation for \( m(t) \) and found \( C \) using initial conditions, you've reached a stage where the equation represents a single specific solution. You start with something like:

• The general solution: \( m(t) = -2000 + Ce^{0.1t} \)
• Substitute \( C \) value back in

Finally, substitute \( C = 3000 \) into the solution, resulting in the particular solution: \( m(t) = -2000 + 3000e^{0.1t} \).

This solution is now not just a formula that could fit many conditions but has been tailored to fit this initial condition, making it a unique depiction of the situation being modeled.