)tan%5E(-1)x-dx%2B2y(1%2Bx%5E(2))dy%3D0.jpeg "(1+y^(2))tan^(-1)x Dx+2y(1+x^(2))dy=0")
Differential Equation: (1+y^(2))tan^(-1)x dx + 2y(1+x^(2))dy = 0
In this article, we will explore the differential equation (1+y^(2))tan^(-1)x dx + 2y(1+x^(2))dy = 0. We will discuss the form of the equation, its classification, and the methodology to solve it.
Form of the Equation
The given differential equation is of the form:
M(x, y)dx + N(x, y)dy = 0
where M(x, y) = (1+y^(2))tan^(-1)x and N(x, y) = 2y(1+x^(2)).
Classification
The differential equation (1+y^(2))tan^(-1)x dx + 2y(1+x^(2))dy = 0 is an exact differential equation.
Methodology to Solve
To solve this differential equation, we can use the method of exact differential equations.
Step 1: Verify the Exactness Condition
First, we need to verify that the equation satisfies the exactness condition, i.e., (∂M/∂y) = (∂N/∂x).
Step 2: Find the Function ψ(x, y)
If the equation is exact, then there exists a function ψ(x, y) such that ∂ψ/∂x = M(x, y) and ∂ψ/∂y = N(x, y).
Step 3: Integrate ψ(x, y)
Integrate ψ(x, y) with respect to x, treating y as a constant, and then integrate with respect to y, treating x as a constant.
Step 4: General Solution
The general solution of the differential equation is ψ(x, y) = C, where C is the constant of integration.
Conclusion
In this article, we discussed the differential equation (1+y^(2))tan^(-1)x dx + 2y(1+x^(2))dy = 0. We classified it as an exact differential equation and outlined the methodology to solve it using the method of exact differential equations. The general solution of the equation is ψ(x, y) = C, where C is the constant of integration.
