MHB Using Euler's Method to find the general solution to the diff. equation

[manzneedshelp]

Could someone please provide guidance on how to begin this problem? I've attached the preface to the assignment question.Show that the general solution of the differential equation
y″(x)=−y(x)
is
y(x)=Acos(x)+Bsin(x)
where A and B are arbitrary constants. Hint: You'll need the Taylor series representations for sin and cos.View attachment 8274

 

[HOI]

"Start" by doing exactly what was done in the excerpt you post! Look for a solution of the form $y= \sum_{n=0}^\infty a_nx^{n}$. Then $y'= \sum_{n=1}^\infty na_nx^{n-1}$ and $y''= \sum_{n= 2}^\infty n(n-1)a_nx^{n-2}$. The differential equation y''= -y becomes $\sum_{n=2}^\infty n(n-1)a_nx^{n-2}= \sum_{n=0}^\infty -a_nx^{n}$. In order to compare them "term by term" we want to adjust the exponent on the left: let j= n- 2.
Then n= j+ 2. When n= 2, j= 0 . Of course "going to infinity" the difference between n and n- 2 doesn't matter. The sum on the left becomes $\sum_{j= 0}^\infty (j+ 2)(j+1)a_{j+ 2}x^{j}$. But the indices in these sums are "dummy indices". The results of the sums won't have an "n" or a "j" in them. We can as easily use "n" instead of "j" on the right. We can write that equation as
$\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n= \sum_{n=0}^\infty -a_n x^n$.

If a two power sums are equal then "corresponding coefficients" are equal- we must have $(n+2)(n+ 1)a_{n+2}= -a_n$ for all n. Of course, that means we must have $a_{n+2}= -\frac{a_n}{(n+2)(n+1)}$.

Now look at the "fundamental solutions" to this equation. For a linear second order equation such as this the "fundamental solutions" are (1) the solution that satisfies y(0)= 0, y'(0)= 1 and (2) the solution that satisfies y(1)= 1, y'(0)= 0. Those will give two independent solutions so that any solution can be written as a linear combination of them.
If y(0)= 0 and y'(0)= 1 then $a_0= 0$ and $a_1= 1$. By the recursion formula, $a_{n+2}= -frac{a_n}{(n+1)(n+2)}$, it is easy to see that $a_2= -\frac{a_0}{1(2)}= 0$, $a_4= -\frac{a_2}{2(3)}= 0$, etc. $a_n= 0$ for all EVEN n. By that same recursion formula, $a_3= -\frac{a_1}{2*3}= \frac{1}{3!}$, $a_5= -\frac{a_3}{4(5)}= -\frac{-11}{2(3)(4)(5)}= \frac{1}{5!}$. Continuing like that it is easy to see that $a_{2n+1}= \frac{(-1)^n}{(2n+1)!}$. So the solution to y'= -y such that y(0)= 0, y'(1)= 1 is $y(x)= \sum_{n=0}^n \frac{(-1)^n}{(2n+1)!}x^{2n+1}$. Now what function has that power series? (Remember that your text specifically said "Hint: You'll need the Taylor series representations for sin and cos.")