Problem 16 - Calculus: Early Transcendentals

Question

Find the solution of the differential equation that satisfies the given initial condition. \( \frac {dP}{dt} = \sqrt {Pt}, P(1) = 2 \)

Step-by-Step Solution

Answer

The solution is \( P(t) = \left( \frac{1}{3}t^{3/2} + \sqrt{2} - \frac{1}{3} \right)^2 \).

Step 1: Separate Variables

The given differential equation is \( \frac{dP}{dt} = \sqrt{Pt} \). We can rewrite this as \( \frac{dP}{\sqrt{P}} = \sqrt{t} \, dt \) by separating the variables \( P \) and \( t \). This is essential to integrate both sides later.

Step 2: Integrate Both Sides

Now integrate both sides of the equation. We have the integral \( \int \frac{1}{\sqrt{P}} \, dP = \int \sqrt{t} \, dt \). The integral on the left is \( 2\sqrt{P} + C_1 \), and on the right it is \( \frac{2}{3}t^{3/2} + C_2 \). We can combine the constants to form a single constant \( C \).

Step 3: Form the General Solution

Equating both integrals, we get \( 2\sqrt{P} = \frac{2}{3}t^{3/2} + C \). This is the general solution of the differential equation. To find \( C \), we'll use the initial condition.

Step 4: Apply Initial Condition

Use the initial condition \( P(1) = 2 \) to find the constant \( C \). Substituting these values into the equation, we get \( 2\sqrt{2} = \frac{2}{3} \cdot 1^{3/2} + C \). Simplifying, \( 2\sqrt{2} = \frac{2}{3} + C \). Solving for \( C \) gives us \( C = 2\sqrt{2} - \frac{2}{3} \).

Step 5: Write the Particular Solution

Substitute \( C \) back into the general solution to obtain the particular solution: \( 2\sqrt{P} = \frac{2}{3}t^{3/2} + 2\sqrt{2} - \frac{2}{3} \). Simplifying further, we have \( \sqrt{P} = \frac{1}{3}t^{3/2} + \sqrt{2} - \frac{1}{3} \). Finally, squaring both sides yields \( P = \left( \frac{1}{3}t^{3/2} + \sqrt{2} - \frac{1}{3} \right)^2 \).

Key Concepts

Separation of Variables

Separation of variables is a common technique used to solve differential equations. Its aim is to rearrange the equation in a way that all terms involving one variable are on one side and all terms involving the other variable are on the opposite side. This simplification helps us prepare for integration, which is the next logical step in solving these equations. In our specific exercise, we start with the differential equation \( \frac{dP}{dt} = \sqrt{Pt} \).

• First, recognize that \( dP \) and \( dt \) need to be separated.
• Reorganize to get \( \frac{dP}{\sqrt{P}} = \sqrt{t} \, dt \).

This rearrangement allows you to integrate each side independently as they're now dependent solely on each respective variable.

Initial Value Problem

An initial value problem (IVP) is a differential equation paired with a value, which typically includes a condition of the form \( P(1) = 2 \). This initial condition is crucial as it allows us to determine the specific solution that passes through this point. Essentially, the IVP refines the general solution down to a unique one.Here's how it works in this exercise:

• Once the general solution of the form \( 2\sqrt{P} = \frac{2}{3}t^{3/2} + C \) is derived, the initial condition \( P(1) = 2 \) helps to find \( C \).
• Substituting \( t = 1 \) and \( P = 2 \), you substitute these into the equation and solve for \( C \).

Without applying the initial condition, the problem lacks the unique solution that fits the criteria given.

Integration

Integration is the mathematical process of finding the integral of functions. It's a key tool in solving differential equations, as it allows us to reverse the operation of differentiation. In the separation of variables method, after isolating the variables, both sides of the equation are integrated.Integrating the equation \( \frac{dP}{\sqrt{P}} = \sqrt{t} \, dt \) gives you:

• Left side: \( \int \frac{1}{\sqrt{P}} \, dP = 2\sqrt{P} + C_1 \).
• Right side: \( \int \sqrt{t} \, dt = \frac{2}{3}t^{3/2} + C_2 \).

It is customary to merge the constants from each side into a single constant \( C \). The act of integration helps bridge between differentiation and the original function's equation.

Particular Solution

The particular solution to a differential equation is the specific solution that satisfies the initial condition. It stands distinct from the general solution, which includes an arbitrary constant. Once this constant is calculated using the initial condition, substituting back provides the particular solution.In this exercise, the steps to reach the particular solution are:

• Start with the general form: \( 2\sqrt{P} = \frac{2}{3}t^{3/2} + 2\sqrt{2} - \frac{2}{3} \).
• Simplify and solve to express \( P \).

Finally, after integrating and finding \( C \), we insert it back, squaring as needed, to achieve the particular solution: \( P = \left( \frac{1}{3}t^{3/2} + \sqrt{2} - \frac{1}{3} \right)^2 \). This solution uniquely fits the given conditions.