(5 + x2)y'' − xy' + 10y = 0, x0 = 0
Find the first four terms in each of two solutions
y1
and
y2
To find the first four terms in each of the two solutions y1 and y2, we can solve the given second-order linear homogeneous differential equation:
(5 + x^2)y'' − xy' + 10y = 0
To solve this equation, let's use the power series method. We assume that the solutions y1 and y2 can be expressed as power series expansions around x = 0. We substitute the power series expansions into the differential equation and equate the coefficients of like powers of x to find the coefficients of the series.
Let's write the power series expansions for y1 and y2 as:
y1 = a0 + a1x + a2x^2 + a3x^3 + ...
y2 = b0 + b1x + b2x^2 + b3x^3 + ...
Differentiating the power series expansions, we get:
y1' = a1 + 2a2x + 3a3x^2 + ...
y2' = b1 + 2b2x + 3b3x^2 + ...
Taking the second derivative, we obtain:
y1'' = 2a2 + 6a3x + ...
y2'' = 2b2 + 6b3x + ...
Next, we substitute these power series expansions and their derivatives into the differential equation:
(5 + x^2)(2a2 + 6a3x + ...) - x(b1 + 2b2x + 3b3x^2 + ...) + 10(a0 + a1x + a2x^2 + a3x^3 + ...) = 0
Now, equate the coefficients of like powers of x to find the coefficients of the series. We will find the coefficients up to the fourth term.
For the term with x^0, we have:
5a0 + 10a0 = 0
15a0 = 0
a0 = 0
Similarly, for x^1, we have:
5a1 + 2a2 - b1 + 10a1 = 0
15a1 + 2a2 - b1 = 0
For x^2, we have:
5a2 + 2a1 + 6a3 - 2b2 + 10a2 = 0
15a2 + 2a1 + 6a3 - 2b2 = 0
For x^3, we have:
5a3 + 3a2 + 6a1 + 10a3 = 0
15a3 + 3a2 + 6a1 = 0
Solving these systems of equations will give us the values of the coefficients a0, a1, a2, a3, b1, b2, b3.
Repeat the process for y2 by substituting its power series expansion into the differential equation and equating coefficients.
By solving these equations algebraically, we can find the first four terms in each of the solutions y1 and y2.