Time remaining:
##### differential equations

 Mathematics Tutor: None Selected Time limit: 1 Day

x2 y''+xy'+(x2 - 1/4)y=0

i shifted the index by 2 and when i moved the other series up i end up with 2 sets of equations containing r and come up with 4 roots for some reason

Apr 2nd, 2015

Let a solution of the differential equation be represented as a series Σn=0 an xn+r , where a0 ≠ 0.

After substitution of the series into the differential equation we get

Σn=0 an [(r+n)2 – 1/4]xn+r + Σn=2 an–2 xn+r  = 0

If n = 0, then we have r2 – 1/4 = 0 and r = 1/2 or r = – 1/2.

If n = 1, then a1 = 0 because (r+1)2 – 1/4 ≠ 0. For n > 1 the coefficients can be found from the relations

an = – an–2 /[(r+n)2 – 1/4] where r = 1/2 or r = – 1/2. Thus we get two linearly independent solutions of the differential equation. Its general solution can be obtained as their linear combination with arbitrary constants C1 and C2.

Apr 3rd, 2015

...
Apr 2nd, 2015
...
Apr 2nd, 2015
Dec 3rd, 2016
check_circle
check_circle
check_circle

Secure Information

Content will be erased after question is completed.

check_circle