How do you use Power Series to solve the differential equation #y'-y=0# ?

Question

How do  you use Power Series to solve the differential equation #y'-y=0# ?

Christopher Allers · Accepted Answer

The solution is #y=c_0sum_{n=0}^infty{x^n}/{n!}=c_0e^x#, where #c_0# is any constant. Let us look at some details. Let 
#y=sum_{n=0}^infty c_n x^n#
#y'=sum_{n=1}^infty nc_n x^{n-1}=sum_{n=0}^infty(n+1)c_{n+1}x^n# So, we can rewrite #y'-y=0# as #sum_{n=0}^infty (n+1)c_{n+1} x^n-sum_{n=0}^infty c_n x^n=0# by combining the summations, #Rightarrow sum_{n=0}^infty[(n+1)c_{n+1}-c_n]x^n=0# so, we have #(n+1)c_{n+1}-c_n=0 Rightarrow c_{n+1}=1/{n+1}c_n# Let us observe the first few terms. #c_1=1/1c_0=1/{1!}c_0# #c_2=1/2c_1=1/{2}cdot1/{1!}c_0=1/{2!}c_0# #c_3=1/3c_2=1/3cdot1/{2!}c_0=1/{3!}c_0#
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#c_n=1/{n!}c_0# Hence, the solution is #y=sum_{n=0}^infty1/{n!}c_0x^n=c_0sum_{n=0}^infty{x^n}/{n!}=c_0e^x#, where #c_0# is any constant.

Elijah Abram · Answer

undefined To solve the differential equation $y' - y = 0$ using power series, you assume a power series solution of the form $y(x) = \sum_{n=0}^{\infty} a_nx^n$. Substitute this series into the differential equation, equate coefficients of like powers of $x$, and solve for the coefficients $a_n$. Then, express the solution in terms of the coefficients.