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find the supply function \(x=f(p)\) that satisfies the initial conditions. $$ \frac{d x}{d p}=\frac{10}{\sqrt{p-3}}, \quad x=100 \text { when } p=\$ 3 $$

Short Answer

Expert verified
The supply function that satisfies the initial conditions is \(x = f(p) = 20 \sqrt{p-3} + 100\).

Step by step solution

01

Integrate the differential equation

Start with the given differential equation \(\frac{d x}{d p}=\frac{10}{\sqrt{p-3}}\). We can write this in the form \(dx = \frac{10}{\sqrt{p-3}} dp\). Integrate both sides to obtain \(x = 20 \sqrt{p-3} + C\), where \(C\) is the constant of integration.
02

Determine the constant of integration

Apply the initial condition \(x=100\) when \(p=3\) to the equation. So, \(100 = 20 \sqrt{3-3} + C \). Hence, we get that the constant \(C = 100\).
03

Final Form of the Supply Function

Insert the constant of integration into the function to get the final form of the supply function, \(x = f(p) = 20 \sqrt{p-3} + 100\)

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Key Concepts

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

Differential Equations
Differential equations are mathematical equations that relate a function with its derivatives. In our problem, we started with the differential equation \(\frac{d x}{d p}=\frac{10}{\sqrt{p-3}}\). This equation connects the rate of change of \(x\) with respect to \(p\). To solve it, we focus on finding the function \(x\) that satisfies this relationship.
Understanding differential equations is crucial because they model how things change and grow. They are used in various fields like physics, engineering, and economics. When dealing with such equations, we often convert them into integrable forms to solve them. In this case, we rearrange it to \(dx = \frac{10}{\sqrt{p-3}} dp\), paving the way for integration.
Integration
Integration is a fundamental concept in calculus that essentially reverses differentiation. When you have a derivative (as in our problem's differential equation), integration helps you find the original function, or at least a family of functions. For solving the given equation, \(\int dx = \int \frac{10}{\sqrt{p-3}} dp\), we integrate on both sides.
The integration results in \(x = 20 \sqrt{p-3} + C\), which indicates that \(x\) is related to \(p\) through the square root function, plus a constant \(C\). This has a crucial role in determining how supply behaves with changing \(p\). The constant \(C\) represents any initial conditions not accounted for in the integration alone, emphasizing the need to consider all initial constraints to fully solve the equations.
Initial Conditions
Initial conditions are necessary to determine the specific solution from a family of solutions given by the general equation\(x = 20 \sqrt{p-3} + C\). These are conditions that allow us to find the particular constant term \(C\) in the integration process. For instance, we used \(x=100\) when \(p=3\) to find \(C\).
This specificity ensures we get the exact solution for the supply function rather than a general one. By substituting these values, we calculated that \(C = 100\). Initial conditions act as the additional information needed to tailor the solution to fit the particular scenario being modeled, thus helping build a complete picture.

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