(1 − 2x − X2)y'' + 2(1 + X)y' − 2y = 0 Y1 = X + 1

Jasonbradley

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Jun 16, 2024

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Solving the Differential Equation (1 − 2x − x2)y'' + 2(1 + x)y' − 2y = 0 with y1 = x + 1

This article explores the solution of the second-order linear homogeneous differential equation:

(1 − 2x − x²)y'' + 2(1 + x)y' − 2y = 0

We are given a particular solution y1 = x + 1. Our goal is to find the general solution of this differential equation.

Method of Reduction of Order

The method of reduction of order is used when we know one solution (y1) to a second-order linear homogeneous differential equation and want to find a second, linearly independent solution (y2).

Here's how the method works:

1. Assume a second solution: We assume a second solution of the form y2 = v(x)y1, where v(x) is an unknown function.

2. Substitute into the differential equation: Substitute y2 and its derivatives into the original differential equation.

3. Simplify and solve for v''(x): The equation will simplify, allowing you to solve for v''(x).

4. Integrate to find v(x): Integrate v''(x) twice to find v(x).

5. Obtain y2: Substitute the found v(x) back into y2 = v(x)y1 to obtain the second solution.

6. General Solution: The general solution is then given by y(x) = c1y1 + c2y2, where c1 and c2 are arbitrary constants.

Applying the Method

Let's apply the method of reduction of order to our problem:

1. Assume a second solution: y2 = v(x)(x + 1)

2. Calculate derivatives:

   • y2' = v'(x)(x + 1) + v(x)
   • y2'' = v''(x)(x + 1) + 2v'(x)

3. Substitute into the differential equation: (1 − 2x − x²)[v''(x)(x + 1) + 2v'(x)] + 2(1 + x)[v'(x)(x + 1) + v(x)] − 2[v(x)(x + 1)] = 0

4. Simplify and solve for v''(x): After simplifying, the equation becomes: (1 − 2x − x²)v''(x) = 0 Therefore, v''(x) = 0

5. Integrate to find v(x): Integrating twice, we get: v'(x) = c1 v(x) = c1x + c2

6. Obtain y2: Substituting v(x) back into y2 = v(x)(x + 1), we get: y2 = (c1x + c2)(x + 1)

7. General Solution: The general solution is given by: y(x) = c1(x + 1) + c2(c1x + c2)(x + 1)

Conclusion

Using the method of reduction of order, we have successfully found the general solution of the differential equation (1 − 2x − x²)y'' + 2(1 + x)y' − 2y = 0 given the particular solution y1 = x + 1. The general solution is:

y(x) = c1(x + 1) + c2(c1x + c2)(x + 1)

This solution captures all possible solutions to the differential equation, encompassing both the given solution (y1) and a second linearly independent solution (y2). The constants c1 and c2 can take on any real values, allowing for a wide range of solutions.