How do you find a Power Series solution of a differential equation?

Answer 1
Let us solve the differential equation #y''=y# by Power Series Method.
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# . . . #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# . . . #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.
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Answer 2

To find a power series solution of a differential equation, follow these steps:

  1. Write the differential equation as a power series.
  2. Expand the functions involved in the equation as power series.
  3. Substitute the power series expansions into the original differential equation.
  4. Equate coefficients of like powers of the independent variable.
  5. Solve the resulting recurrence relations to find the coefficients of the power series solution.
  6. Write the power series solution with the determined coefficients.
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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