Solving an ODE Using the Exact Method: A Step-by-Step Guide
When solving differential equations (ODEs), the exact method is a powerful tool. This article provides a comprehensive guide to solving the ODE [frac{y^2}{x - y^2} , dx frac{x^2}{x - y^2} , dy 0]. We will break down the process into several steps, including checking for exactness and performing integration.
Identifying the Problem
We start with the given ODE:
[frac{y^2}{x - y^2} , dx frac{x^2}{x - y^2} , dy 0]
Let's denote:
[M frac{y^2}{x - y^2}] [N frac{x^2}{x - y^2}]Checking for Exactness
To determine if the ODE is exact, we need to calculate the partial derivatives [M_y] and [N_x].
Calculating [M_y]
[M_y frac{partial}{partial y} left(frac{y^2}{x - y^2}right)]
Using the quotient rule:
[M_y frac{2xy^2 - 2y(x - y^2)}{(x - y^2)^2} frac{2xy^2 - 2xy 2y^3}{(x - y^2)^2} frac{2y^3 2xy(y - 1)}{(x - y^2)^2}]
Simplifying the expression:
[M_y frac{2y(x^2y - xy^2)}{(x - y^2)^2}]
Calculating [N_x]
[N_x frac{partial}{partial x} left(frac{x^2}{x - y^2}right)]
Using the quotient rule again:
[N_x frac{2x(x - y^2) - x^2}{(x - y^2)^2} frac{2x^2 - 2xy^2 - x^2}{(x - y^2)^2} frac{x^2 - 2xy^2}{(x - y^2)^2}]
Comparing [M_y] and [N_x]:
[M_y eq N_x]
Since [M_y eq N_x], the ODE is not exact.
Multiplying by an Integrating Factor
To make the ODE exact, we need to multiply by an integrating factor. Let's try [e x - y^2] as suggested.
[frac{y^2}{e} , dx frac{x^2}{e} , dy 0]
Multiplying both sides by [frac{1}{x^2y^2}] gives:
[frac{1}{x^2y^2} cdot frac{y^2}{x - y^2} , dx frac{1}{x^2y^2} cdot frac{x^2}{x - y^2} , dy 0]
Simplifying:
[left(frac{1}{x^2} cdot frac{1}{x - y^2} right) , dx left(frac{1}{y^2} cdot frac{1}{x - y^2} right) , dy 0]
The equation is now exact:
[frac{1}{x^2} , dx frac{1}{y^2} , dy 0]
Integrating both terms:
[-frac{1}{x} - frac{1}{y} -C]
[frac{1}{x} frac{1}{y} C]
Where [C] is an arbitrary constant.
Conclusion
By following these steps, we have solved the ODE [frac{y^2}{x - y^2} , dx frac{x^2}{x - y^2} , dy 0]. The final solution is given by:
[frac{1}{x} frac{1}{y} C]
The use of an integrating factor and the exact method allowed us to solve the ODE efficiently. This method can be applied to other differential equations, making it a valuable skill in the study of ODEs.