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

In summary, the function that has the power series with coefficients that equal the negative of the sum of the squares of the integers from 0 to n-1 is the function that represents the inverse cosine of the given number.
  • #1
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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
 

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"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.")
 

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

1. What is Euler's Method?

Euler's Method is a numerical technique used to approximate the solutions to a differential equation. It uses small increments to estimate the values of the solution at different points.

2. How does Euler's Method work?

Euler's Method works by using the initial condition of a differential equation to find the slope of the tangent line at that point. It then uses this slope to estimate the next point on the graph, and repeats this process to approximate the solution at other points.

3. What is the general solution to a differential equation?

The general solution to a differential equation is an equation that satisfies the given differential equation for all possible solutions. It includes a constant term that accounts for any variation in the solutions.

4. When is Euler's Method useful?

Euler's Method is most useful when analytical solutions to a differential equation are difficult or impossible to find. It is also useful for solving systems of differential equations and for visualizing the behavior of a solution over time.

5. What are some limitations of Euler's Method?

Euler's Method is a first-order approximation, meaning that it can introduce errors when used to approximate complex or rapidly changing solutions. It also requires small increments to accurately estimate the solution, which can be computationally expensive.

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