Answer:
The given power series [tex]y =\sum^{\infty}_{n=0} {(-1)^n x^{2n}}[/tex] is a solution of the differential equation (1+x^2)y' + 2xy = 0
Step-by-step explanation:
This is a very trivial exercise, follow the steps below for the solution:
Step 1: Since n = 0, 1, 2, 3, 4, ........, Substitute the values of n into equation (1) below.
[tex]y =\sum^{\infty}_{n=0} {(-1)^n x^{2n}}[/tex].....................(1)
[tex]y = 1 - x^2 + x^4 - x^6 + x^8.........[/tex]
Step 2: Find the derivative of y, i.e. y'
[tex]y' = -2x + 4x^3 - 6x^5 + 8x^7 .............[/tex]
Step 3: Substitute y and y' into equation (2) below:
[tex](1+x^2)y' + 2xy = 0\\\\(1+x^2)(-2x + 4x^3 - 6x^5 + 8x^7......) + 2x(1 - x^2 + x^4 - x^6 + x^8.......) = 0\\\\-2x+ 4x^3 - 6x^5 + 8x^7........ - 2x^3 +4x^5 - 6x^7 + 8x^9 ......+ 2x - 2x^3 + 2x^5 - 2x^7 + 2x^9...... = 0\\\\0 = 0[/tex]
(Verified)
Since the LHS = RHS = 0, the given power series [tex]y =\sum^{\infty}_{n=0} {(-1)^n x^{2n}}[/tex] is a solution of the differential equation (1+x^2)y' + 2xy = 0
