Introduction to the Problem
The core of this discussion revolves around a fundamental problem in calculus: given a point (1, 1) and a slope (frac{dy}{dx} frac{2y}{x}), we aim to find the curve that satisfies both the equation of the slope and the condition of passing through the specified point. Understanding this process will enhance one's grasp of differential equations and their applications.
Solving the First-Order Differential Equation
The first step in solving this problem is to recognize it as a first-order differential equation. Let's break down the solution into manageable steps.
Separation of Variables
Given the equation (frac{dy}{dx} frac{2y}{x}), we rearrange it to separate the variables. This yields:
(frac{dy}{y} frac{2}{x} dx)
Integration
Integrating both sides of the equation, we obtain:
(int frac{dy}{y} int frac{2}{x} dx)
Which simplifies to:
(ln|y| 2ln|x| C)
Here, C is the constant of integration.
Exponentiation and Simplification
To remove the logarithms, we exponentiate both sides:
(|y| e^{2ln|x| C} e^C cdot e^{2ln|x|} e^C cdot |x|^2 kx^2)
Where we define k e^C as a new constant.
Hence, the general form of the curve is:
(y kx^2)
Using the Initial Condition
Finally, we use the initial condition that the curve passes through the point (1, 1) to find the value of k. Substituting x 1 and y 1 into the equation, we get:
(1 k(1)^2 Rightarrow k 1)
Thus, the specific equation of the curve is:
(y x^2)
Conclusion
In summary, the curve that passes through the point (1, 1) and has a slope given by the equation (frac{dy}{dx} frac{2y}{x}) is found to be:
(y x^2)
Additional Insights
This solution method is a classic example of using separation of variables to solve a first-order differential equation. It demonstrates how the integration of logarithmic functions and the application of initial conditions can lead to a specific curve. This approach is not only useful in pure mathematics but also in various scientific and engineering applications, where understanding the behavior of curves based on given conditions is crucial.