# Y'' + y = 0 solution and recursion relation

• I
• Sam D
In summary, the general solution for the equation is y(x) = C1cos(x) + C2sin(x) and the recursion relation for the equation is An+2 = -An / (n+2)(n+1). To show that the recursion relation is equivalent to the general solution, one can take the sum of the power series of sine and cosine multiplied by a constant and show that the result satisfies both the differential equation and the recurrence relation. This can be done by calculating the first few terms of the series in terms of A0 and A1 and recognizing the resulting series for even and odd n values. The final solution is y(x) = A0cos(x) + A1sin(x), where A0
Sam D
I've found the general solution to be y(x) = C1cos(x) + C2sin(x).

I've also found a recursion relation for the equation to be:

An+2 = -An / (n+2)(n+1)

I now need to show that this recursion relation is equivalent to the general solution. How do I go about doing this?

Any help would be greatly appreciated!

So is this the recurrence relation for the coefficients ##c_i## in the power series representing the solution: ##y(x)=c_0 + c_1 x + c_2 x^2 \dots## ? It shouldn't be difficult to take the sum of the power series of sine and cosine multiplied by a constant, and then show that the result satisfies both the DE and that recurrence relation.

Try calculating the first few terms of the series in terms of ##a_0## and ##a_1##. Hopefully, you'll recognize the resulting two series.

Sam D
<Moderator's note: Approved as it is more than two weeks since the OP has been seen. Member has been warned not to post full solutions. This is an exception as it closes the thread.>

Let's start from your reccurence relation:

##A_{n+2}=\frac{-A_N}{(n+2)(n+1)}##

First collect even ##n## values:

##A_2= \frac{-A_0}{2.1}##

##A_4= \frac{-A_2}{4.3}=\frac{A_0}{4.3.2.1}=\frac{A_0}{4!}##

##A_6= \frac{-A_4}{6.5}=\frac{-A_0}{6.5.4.3.2.1}=\frac{-A_0}{6!}##

...

Now take odd n values:

##A_3= \frac{-A_1}{3.2}##;

##A_5= \frac{-A_3}{5.4}=\frac{A_1}{5.4.3.2.1}=\frac{A_0}{5!}##

##A_7= \frac{-A_5}{7.6}=\frac{-A_0}{7.6.5.4.3.2.1}=\frac{-A_0}{7!}##

...

So final solution isOn putting the values of ##A_n## in Maclaurin series solution (##y(x)=\sum_{n=0}{ A_n x^n}##),##y(x)=\sum_{n=0}{ \frac{(-1)^n x^{2n}}{ 2n!} + \frac{(-1)^n x^{2n+1}}{ 2n+1!}}##

##y(x)= A_0 \cos (x) + A_1 \sin (x)##

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## 1. What is the solution to the equation "Y'' + y = 0"?

The solution to this equation is a sinusoidal function, specifically of the form y = A*sin(x) + B*cos(x), where A and B are constants determined by the initial conditions of the problem.

## 2. How does the recursion relation impact the solution?

The recursion relation determines the values of the constants A and B in the solution. It relates the current value of the function with its previous values, and can be used to find the values of A and B for any given initial condition.

## 3. Can you explain the meaning of the second derivative in this equation?

The second derivative represents the rate of change of the first derivative. In this equation, it represents the acceleration of the function, which is equal to the negative of the function itself.

## 4. What are some real-world applications of this equation and its solution?

This equation and its solution are used in many fields, including physics, engineering, and finance. It can be used to model oscillatory motion, such as the motion of a pendulum or a spring, and can also be applied to analyze the behavior of electrical circuits and financial markets.

## 5. How does the solution to this equation change if there are external forces acting on the system?

If there are external forces acting on the system, the equation would become "Y'' + y = F(x)", where F(x) represents the external force function. The solution would then be y = A*sin(x) + B*cos(x) + F(x), where A and B are constants determined by the initial conditions and F(x) is the particular solution for the external force.

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