In each of Problems 1 through 11: a. Seek power series...

In each of Problems 1 through 11: a. Seek power series solutions of the given differential equation about the given point x0; find the recurrence relation that the coefficients must satisfy. b. Find the first four nonzero terms in each of two solutions y1 and y2 (unless the series terminates sooner). c. By evaluating the Wronskian W[y1, y2](x0), show that y1 and y2 form a fundamental set of solutions. d. If possible, find the general term in each solution. 1. y'' - y = 0, x0 = 0 2. y'' + 3y' = 0, x0 = 0 3. y'' - xy' - y = 0, x0 = 0 4. y'' - xy' - y = 0, x0 = 1 5. y'' + k^2x^2y = 0, x0 = 0, k a constant 6. (1 - x)y'' + y = 0, x0 = 0

Transcript

00:01 In first a part, the given differential equation y double dash, negative y is equal to 0, given x naught is equal to 0.

00:09 So now in first a part to find power series solution of differential equation.

00:24 So let y of x summation vary from n is equal to 0 to infinity a n x to the power of n be a solution.

00:41 So now differentiate the y with respect to x then we have summation vary from n equal to 0 to infinity a n multiplied to n x to the power of n negative 1.

00:54 Again differentiate then we have summation vary from n equal to 0 to infinity n multiplied to n negative 1 a n x to the power of n negative 2.

01:06 So this quantity in a summation vary from n equal to 0 to infinity n plus 1 multiplied to n plus 2 a n plus 2 x to the power of n.

01:20 Put this y double dash and y in given power series then we have summation n equal to 0 to infinity a n plus 2 multiplied to n plus 1 multiplied to n plus 2 x to the power negative summation vary from 0 to infinity a n x to the power n is equal to 0.

01:43 So now from here it implies summation vary from n equal to 0 to infinity.

01:49 So we have n plus 1 multiplied to n plus 2 a n plus 2 negative a n multiplied to x to the power of n equal to 0.

02:01 So from here n plus 1 multiplied to n plus 2 a n plus 2 is equal to a n.

02:11 So now for n equal to 0 then we have 1 multiplied to that is a 2 is equal to a naught.

02:21 So from here value of a 2 is equal to a naught upon 1 multiplied to that is a naught over 2 factorial.

02:29 Now for n is equal to 1 then we have 2 multiplied to 3 that is a 3 is equal to a 1...

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