We seek the solution of the equation in the form of a power series
y(x)=∑n=0∞anxn Then ODE:
(x2−1)n=2∑∞ann(n−1)xn−2+3xn=1∑∞annxn−1+xn=0∑∞anxn=0n=2∑∞ann(n−1)xn−2a2−6a3x−n=2∑∞an+2(n+2)(n+1)xn+3a1x+3n=2∑∞annxn+a0x+n=2∑∞an−1xn=0
⎩⎨⎧−2a2=0−6a3+3a1+a0=0ann(n−1)−an+2(n+2)(n+1)+3ann+an−1=0,n=2,3,4...⎩⎨⎧a2=0a3=63a1+a0an+2=(n+2)(n+1)ann(n+2)+an−1,n=2,3,4... Answer:
y(x)=a0+a1x+63a1+a0x3+n=2∑∞(n+2)(n+1)ann(n+2)+an−1xn+2
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