y%26%2339%3B-xy%3Dx%5E3-x.jpeg "(x^2-1)y'-xy=x^3-x")
Solving the First-Order Linear Differential Equation: (x^2-1)y' - xy = x^3 - x
This article will explore the solution of the first-order linear differential equation:
(x^2-1)y' - xy = x^3 - x
1. Standard Form
First, we need to rewrite the equation in standard form, which is:
y' + p(x)y = q(x)
Dividing both sides of the given equation by (x^2-1), we get:
y' - (x/(x^2-1))y = (x^3 - x)/(x^2-1)
Therefore, we have:
- p(x) = -x/(x^2-1)
- q(x) = (x^3 - x)/(x^2-1)
2. Integrating Factor
The integrating factor is given by:
μ(x) = exp(∫p(x) dx)
Let's calculate the integrating factor:
μ(x) = exp(∫-x/(x^2-1) dx)
This integral can be solved using a substitution. Let u = x^2 - 1, then du = 2x dx.
μ(x) = exp(∫-1/(2u) du)
μ(x) = exp(-1/2 ln|u|)
μ(x) = exp(ln|u|^(-1/2))
μ(x) = |u|^(-1/2) = 1/√(x^2-1)
3. Solving the Equation
Multiplying both sides of the standard form equation by the integrating factor, we obtain:
(1/√(x^2-1))y' - (x/(√(x^2-1)(x^2-1)))y = (x^3 - x)/(√(x^2-1)(x^2-1))
The left-hand side is now the derivative of a product:
d/dx [(1/√(x^2-1))y] = (x^3 - x)/(√(x^2-1)(x^2-1))
Integrating both sides with respect to x, we have:
(1/√(x^2-1))y = ∫(x^3 - x)/(√(x^2-1)(x^2-1)) dx + C
The integral on the right-hand side can be solved using partial fractions or other integration techniques. The solution will involve a combination of logarithmic and inverse trigonometric functions.
4. General Solution
Finally, multiplying both sides by √(x^2-1) gives us the general solution for y:
y = √(x^2-1) [∫(x^3 - x)/(√(x^2-1)(x^2-1)) dx + C]
Conclusion
The solution to the given differential equation involves finding the integrating factor, multiplying the equation by it, integrating both sides, and solving for y. The solution will involve a constant of integration (C), leading to a general solution that represents a family of curves.