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Find a power series of the form

$$u=\sum_{k=0}^{\infty }a_kx^k$$

for the D.E. as follows with the initial conditions

$$u''+x^2u=0$$$$u(0)=1,u'(0)=0$$$$u=\sum_{n=0}^{\infty }a_nx^n$$$$u'=\sum_{n=0}^{\infty }na_nx^{n-1}=\sum_{n=0}^{\infty }(n+1)a_{n+1}x^n$$$$u''=\sum_{n=0}^{\infty }n(n-1)a_nx^{n-2}=\sum_{n=0}^{\infty }n(n+1)a_{n+1}x^n$$$$x^2u=\sum_{n=0}^{\infty }a_nx^{n+2}=\sum_{j=2}^{\infty }a_{j-2}x^j$$$$j=n+2,n=j-2$$$$=\sum_{n=2}^{\infty }a_{n-2}x^n=\sum_{n=0}^{\infty }a_{n-2}x^n$$$$a_{-2}=0, a_{-1}=0$$

But where to from here?

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If I may suggest, do not shift the index $$u''+x^2 u=0\implies \sum_{n=0}^\infty n(n-1)a_nx^{n-2}+\sum_{n=0}^\infty a_nx^{n+2}=0$$ So, for the same degree $k$ we have $$(k+1)(k+2)a_{k+2}+a_{k-2}=0$$ that is to say $$a_{k+2}=-\frac{a_{k-2} }{(k+1)(k+2) }$$ Now $$u(0)=1\implies a_0=1$$ $$u'(0)=0\implies a_1=0$$ Looking at the first terms, we have

$$u''+x^2u=2 a_2+6 a_3 x+(a_0+12 a_4) x^2+(a_1+20 a_5) x^3+\cdots$$ which make $a_2=a_3=0$.

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