Problem 14 - Calculus

Question

Find the particular solution to the differential equation, given the general solution and an initial condition. ( \(C\) is the constant of integration.) $$Q=1 /(C t+C) ; Q=4 \text { when } t=2$$

Step-by-Step Solution

Answer

The particular solution is \( Q = \frac{1}{\left(\frac{-2+\sqrt{5}}{2}(t+\frac{-2+\sqrt{5}}{2}) \right)} \).

Step 1: Understand the Given Information

We are given the general solution: \( Q = \frac{1}{C(t+C)} \). We need to find the value of \( C \) such that the initial condition \( Q = 4 \) when \( t = 2 \) is satisfied.

Step 2: Substitute Initial Condition into the General Solution

Substitute \( t = 2 \) and \( Q = 4 \) into the general solution: \[ 4 = \frac{1}{C(2+C)} \].

Step 3: Solve for the Constant \( C \)

From \( 4 = \frac{1}{C(2+C)} \), rearrange to get: \( C(2+C) = \frac{1}{4} \).Multiply through by \( 4 \) to eliminate the fraction: \( 4C(2+C) = 1 \). Expanding this gives: \( 8C + 4C^2 = 1 \).

Step 4: Rearrange and Solve Quadratic Equation

Write the quadratic equation in standard form: \( 4C^2 + 8C - 1 = 0 \).Use the quadratic formula: \( C = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = 8 \), \( c = -1 \).First, calculate the discriminant: \( b^2 - 4ac = 64 + 16 = 80 \).Then, \( C = \frac{-8 \pm \sqrt{80}}{8} \).Simplifying, \( C = \frac{-8 \pm 4\sqrt{5}}{8} \), which reduces to \( C = \frac{-2 \pm \sqrt{5}}{2} \).

Step 5: Choose the Appropriate Solution

We need to choose the value of \( C \) that creates a valid function for \( Q \) when \( t = 2 \) and \( Q = 4 \). Let's test \( C = \frac{-2 + \sqrt{5}}{2} \).Verify with the original equation: calculate and check if this satisfies the initial condition. Substituting, if valid, that's our solution for \( C \).

Step 6: Write the Particular Solution

Using \( C = \frac{-2 + \sqrt{5}}{2} \), substitute into the general solution to provide the particular solution.

Key Concepts

Particular Solution

In differential equations, a particular solution refers to a specific solution that satisfies both the differential equation and any given initial conditions. Unlike the general solution, which contains arbitrary constants, the particular solution is more refined and complete.
When given a problem involving finding a particular solution, you start with a general solution. The goal is to identify specific values for the constants by utilizing the provided initial conditions.

• Start with the general differential equation solution.
• Apply the initial conditions to solve for any unknown constants.
• The result you get after substituting this back into the original equation is the particular solution.

The particular solution is then a tailored version of the general solution for your specific problem.

Initial Condition

Initial conditions are specific values given for the variables in a differential equation that help in finding the particular solution. They provide a starting point or specific case for the problem.
For instance, in the given exercise, the initial condition is that when the time ( $t$) is 2, the quantity ( $Q$) is 4.

• The initial condition allows you to substitute these known values into your general solution.
• By using the initial condition, you can calculate the unknown constants precisely.
• This process involves solving one or more equations to isolate and determine these constants.

Understanding and correctly applying initial conditions is crucial to narrowing down from a broad general solution to the particular case that fits your problem.

Constant of Integration

The constant of integration is an arbitrary constant that appears when you integrate a function, due to the nature of indefinite integrals. As integration is the reverse of differentiation, and constants vanish when differentiated, we have to reintroduce them as constants of integration:

• When a differential equation is solved, the result includes a constant of integration because multiple functions could have the same derivative.
• In any particular problem, we use initial conditions to find a specific value for this constant so that the solution fits the real-world scenario.
• The constant of integration can often be denoted as $C$, but it can vary.

In the context of our exercise, $C$ was initially arbitrary, and its specific value was derived by using and solving the initial condition provided.

Quadratic Formula

The quadratic formula is a method for solving quadratic equations, which are polynomial equations of degree two. The general quadratic equation is expressed as \(ax^2 + bx + c = 0\). The quadratic formula to determine the roots (solutions) of a quadratic equation is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here's how it helps in solving problems:

• Identify the coefficients (\(a\), \(b\), \(c\)) from your equation.
• Calculate the discriminant \(b^2 - 4ac\). If the discriminant is positive, you have two distinct roots; zero gives you one repeated root; negative implies complex roots.
• Use the formula to calculate the roots, which might provide possible solutions to your problem.

In the exercise provided, the quadratic formula was used to solve for the constant \(C\), giving possible values from which the right one was selected to satisfy the initial condition. It's a foundational tool for students dealing with quadratic equations in calculus or algebra.