Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Assume $y=\sum a_n x^n$. The ODE is $$y'' + (2 - 4x^2)y = 0$$

$y(0) = 1, y'(0) = 0$

$a_0 = 1, a_1 = 0, a_2 = -1$

share|improve this question
2  
What have you done? –  Pragabhava Nov 8 '12 at 0:24

2 Answers 2

First note that you can obtain $a_0$ and $a_1$ without going to the ODE itself. Note that since $y(x) = \displaystyle \sum_{n=0}^{\infty} a_n x^n$, $y(0)= a_0 = 1$ and $y'(x) = \displaystyle \sum_{n=0}^{\infty} na_n x^{n-1} = \sum_{n=1}^{\infty} na_n x^{n-1} \implies y'(0) = a_1 = 0$.

Further, $$y''(x) = \displaystyle \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} = \displaystyle \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} = \displaystyle \sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2} x^{n}$$ Hence, the ODE becomes $$\displaystyle \sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2} x^{n} + (2-4x^2) \displaystyle \sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} \left((n+2)(n+1)a_{n+2} x^{n} + (2-4x^2)a_n x^n \right)$$ This gives us $$\sum_{n=0}^{\infty} \left((n+2)(n+1)a_{n+2} x^{n} + 2 a_n x^n -4a_n x^{n+2} \right) = 0$$ i.e. $$(2a_2 + 2a_0) + (6a_3+2a_1)x + \sum_{n=2}^{\infty} \left((n+2)(n+1)a_{n+2} x^{n} + 2 a_n x^n -4a_{n-2} x^{n} \right) = 0$$ Hence, we have the following \begin{align} a_0 & = 1\\ a_1 & = 0\\ a_2 & = - a_0 = -1\\ a_3 & = - \dfrac{a_1}3 = 0\\ a_{n} & = \dfrac{4a_{n-4} - 2a_{n-2}}{n(n-1)} & \forall n \geq 4 \end{align} This gives us that all the odd coefficients are $0$ and only the even coefficients remain. Hence, we have that $a_0 = 1$, $a_2 = -1$ and in general $$a_{2n} = \dfrac{4a_{2n-4} - 2 a_{2n-2}}{2n(2n-1)} \,\,\,\,\,\,\,\, \forall n \geq 2$$

share|improve this answer

To solve the differential equation using the power series technique, you assume the solution to have the form

$$ y(x)=a_o + a_1 x + a_2 x^2+ \dots \,, $$

where $a_0,a_1,a_2,\dots$ are constants to be determined. On the other hand, the Taylor series of the function $ y(x) $ at the point $x=0$ is given by

$$ y(x)=y(0) + \frac{y'(0)}{1!} x + \frac{y''(0)}{2!} x^2+ \dots \,. $$

Now, compare the coefficients of powers of $x$, one can see that

$$ a_0=y(0), \quad a_1 = y'(0), \quad a_2 = \frac{y''(0)}{2!}, \dots, a_n = \frac{y^{(n)}(0)}{n!} \,.$$

Using the given differential equation, you can find the $a_n$, by finding the $y^{(n)}(x)$ by successive differentiation of the differential equation. We find $a_2$ as

$$y''(x)= - (2 - 4x^2)y(x) \implies y''(0)=-(2-0)y(0) = -2$$

$$ \implies a_2 = \frac{y''(0)}{2!}=\frac{-2}{2}=-1\,. $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.