By Power Series Method, we can find
#y=c_0coshx+c_1sinhx#,
where #c_0# and #c_1# are any constants.
Let us look at some details.
Let #y=sum_{n=0}^inftyc_n x^n#, where #c_n# is to be determined.
By taking derivatives term by term,
#y'=sum_{n=1}^{infty}nc_nx^{n-1}#
and
#y''=sum_{n=2}^infty n(n-1)c_nx^{n-2}#
So, #y''=y# becomes
#sum_{n=2}^infty n(n-1)c_nx^{n-2}=sum_{n=0}^inftyc_n x^n#
by shifting the indices on the summation on the left by 2,
#sum_{n=0}^infty(n+2)(n+1)c_{n+2}x^n=sum_{n=0}^inftyc_n x^n#
By matching each coefficients,
#(n+2)(n+1)c_{n+2}=c_n
Rightarrow c_{n+2}=c_n/{(n+2)(n+1)}#
Let us observe the first few even terms,
#c_2=1/{2cdot1}c_0=1/{2!}c_0#
#c_4=1/{4cdot3}c_2=1/{4cdot3}cdot1/{2!}c_0=1/{4!}c_0#
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#c_{2n}=1/{(2n)!}c_0#
Let us observe the first few odd terms,
#c_3=1/{3cdot2}c_1=1/{3!}c_1#
#c_5=1/{5cdot4}c_3=1/{5cdot4}cdot1/{3!}c_1=1/{5!}c_1#
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#c_{2n+1}=1/{(2n+1)!}c_1#
Now, we can find the solution #y#.
#y=sum_{n=0}^infty c_nx^n#
by separating even terms and odd terms,
#=sum_{n=0}^inftyc_{2n}x^{2n}+sum_{n=0}^inftyc_{2n+1}x^{2n+1}#
by using the formulas for #c_{2n}# and #c_{2n+1}# above,
#=c_0sum_{n=0}^inftyx^{2n}/{(2n)!}+c_1sum_{n=0}^infty x^{2n+1}/{(2n+1)!}#
Recall:
#coshx=sum_{n=0}^infty x^{2n}/{(2n)!}#
#sinhx=sum_{n=0}^infty x^{2n+1}/{(2n+1)!}#
Hence, #y=c_0coshx+c_1sinhx#, where #c_0# and #c_1# are any constants.
