y%26%2339%3B%26%2339%3B%2By%3D0.jpeg "(1-x)y''+y=0")
Solving the Differential Equation (1-x)y'' + y = 0
This article will explore the solution to the second-order linear differential equation:
(1-x)y'' + y = 0
This equation is a classic example of a differential equation with a regular singular point at x = 1. Let's break down the process of finding its solution.
Understanding the Equation
- Second-order: The highest derivative in the equation is the second derivative, y''.
- Linear: The equation is linear because the dependent variable (y) and its derivatives appear only to the first power.
- Regular singular point: At x = 1, the coefficient of y'' becomes zero. This point is called a regular singular point because the other coefficient (1) is non-zero.
Using the Frobenius Method
To solve this type of equation, we'll employ the Frobenius method. This method involves finding a series solution of the form:
y(x) = Σ (a<sub>n</sub> x<sup>n+r</sup>)
where:
- r is a constant (the indicial root)
- a<sub>n</sub> are coefficients to be determined
1. Finding the Indicial Equation
We begin by substituting the Frobenius series into the differential equation. After some algebraic manipulation and equating coefficients to zero, we obtain the indicial equation:
r(r-1) = 0
This equation has roots r<sub>1</sub> = 1 and r<sub>2</sub> = 0.
2. Finding the Recurrence Relation
Next, we substitute the series solution and the roots back into the differential equation. This leads to a recurrence relation that defines the coefficients a<sub>n</sub>:
a<sub>n+1</sub> = (n+r)(n+r-1) a<sub>n</sub> / (n+r+1)(n+r)
3. Solving for the Coefficients
We use the recurrence relation to find the values of a<sub>n</sub>. The specific values will depend on the chosen root (r<sub>1</sub> or r<sub>2</sub>).
-
Case 1: r = 1
The recurrence relation becomes:
a<sub>n+1</sub> = (n+1)n a<sub>n</sub> / (n+2)(n+1)
Solving this, we get:
a<sub>n</sub> = a<sub>0</sub> / (n+1)
-
Case 2: r = 0
The recurrence relation becomes:
a<sub>n+1</sub> = n(n-1) a<sub>n</sub> / (n+1)n
Solving this, we get:
a<sub>n</sub> = a<sub>0</sub> / (n+1)
4. Constructing the Solution
Using the coefficients obtained, we can construct the general solution:
y(x) = c<sub>1</sub> y<sub>1</sub>(x) + c<sub>2</sub> y<sub>2</sub>(x)
where:
- y<sub>1</sub>(x) = a<sub>0</sub> Σ (x<sup>n+1</sup> / (n+1))
- y<sub>2</sub>(x) = a<sub>0</sub> Σ (x<sup>n</sup> / (n+1))
c<sub>1</sub> and c<sub>2</sub> are arbitrary constants.
Conclusion
The Frobenius method allows us to find a series solution to the differential equation (1-x)y'' + y = 0. This solution is expressed as a linear combination of two linearly independent series. The specific form of the solution depends on the initial conditions. This example showcases the power of the Frobenius method in solving differential equations with regular singular points.