Thread: Consider the following. y'' − xy' − y = 0

1. Consider the following. y'' − xy' − y = 0

Consider the following.y''xy'y = 0, y(0) = 4, y'(0) = 1

(a) Find the first five nonzero terms in the solution of the given initial value problem.

2. Re: Consider the following. y'' − xy' − y = 0

How far has you been able to get?

$\displaystyle y = \sum_{n = 0}^{\infty}a_n x^n$

$\displaystyle y' = \sum_{n = 1}^{\infty}n~a_n x^{n-1}$

$\displaystyle y'' = \sum_{n = 2}^{\infty}n(n - 1)~a_n x^{n-2}$

$\displaystyle \sum_{n = 2}^{\infty}n(n - 1)~a_n x^{n-2} - x \cdot \sum_{n = 1}^{\infty}n~a_n x^{n-1} - \sum_{n = 0}^{\infty}a_n x^n = 0$

-Dan

3. Re: Consider the following. y'' − xy' − y = 0

I would note that the (MacLaurin) series expansion of the solution $y(x)$ about $x=0$ is
$$y(x) = y(0) + {y'(0) \over 1!}x + {y''(0) \over 2!}x^2 + \cdots + {y^{(n)}(0) \over n!}x^n + \cdots$$
The initial conditions give us $y(0)$ and $y'(0)$ and the equation gives us
$$y''(0) = 0\cdot y'(0) + y(0) = y(0)$$
We can then differentiate (implicitly) the original equation to get
$$y''' - xy'' - 2y' = 0 \implies y'''(0) = 0\cdot y''(0) + 2y'(0)$$
And repeat...

4. Re: Consider the following. y'' − xy' − y = 0

hc23881, is there a reason why you made no attempt to do the problem yourself? If you know the basic definitions, this problem is a little tedious but straightforward.

5. Re: Consider the following. y'' − xy' − y = 0

I know those are teh solutions but im not sure about using the given initial values..? does y(o) equal to a0? and y'(o) = a1?

edit:i was able to get the answer

6. Re: Consider the following. y'' − xy' − y = 0

So this is just a question of arithmetic? If $\displaystyle y(x)= a_0+ a_1x+ a_2x^2+ a_3x^3+ \cdot\cdot\cdot$ where all of the terms in the "$\displaystyle \cdot\cdot\cdot$" involve positive powers of x. So $\displaystyle y(0)= a_0+ a_1(0)+ a_2(0)^2+ a_3(0)^3+ \cdot\cdot\cdot$. That is $a_0= 0$. Similarly, differentiating that sum (it is a power series so differentiable term by term), we get $\displaystyle y'(x)= a_1+ 2a_2x+ 3a_3x^2+ 4a_4x^3+ \cdot\cdot\cdot$. So $\displaystyle y'(0)= a_1+ a_2(0)+ a_3(0)^2+ a_4(0)^3+ \cdot\cdot\cdot$.

7. Re: Consider the following. y'' − xy' − y = 0

Thanks I got that one! I am now having trouble on part of this one
(5 + x2)y'' − xy' + 10y = 0, x0 = 0

I was able to find the recurrence relation: an+2 = -(n^2 -2n+10)/(5(n+2)(n+1))
but I cant get this part
Find the first four terms in each of two solutions
y1
and
y2

8. Re: Consider the following. y'' − xy' − y = 0

$x_0 = 0$? That's just shorthand to say that we want the series expansions about $x=0$?

9. Re: Consider the following. y'' − xy' − y = 0

$x_0 = 0$? That's just shorthand to say that we want the series expansions about $x=0$?

You don't need a recurrence relation (which that isn't because you only have $a_{n+2}$).

Instead look to find expressions for $a_1$ and $a_2$ in terms of $a_0$ and $a_1$.

10. Re: Consider the following. y'' − xy' − y = 0

Originally Posted by hc23881
Thanks I got that one! I am now having trouble on part of this one
(5 + x2)y'' − xy' + 10y = 0, x0 = 0

I was able to find the recurrence relation: an+2 = -(n^2 -2n+10)/(5(n+2)(n+1))
This is NOT a "recurrence relation". Did you mean an+2 = -(n^2 -2n+10)/(5(n+2)(n+1))an?

but I cant get this part
Find the first four terms in each of two solutions
y1

y2
The simplest way to handle this is to assume y1(0)= 1, y1'(0)= 0 and y2(0)= 0, y2(0)= 1.
That will guarantee two independent solutions.

Since y(0)= a0 and y'(0)= a1, you can see immediately that, for y1, a0= 1 and a1= 0. For y2, a0= 0 and a1= 1. Further, since (assuming I was correct above) each an is a multiple of an-2, it is clear that, for y1, all an with n odd is 0, and, for y2, all n with n odd is 0. y1 will be a sum with only even powers of x and y2 will be a sum with only odd powers of x.

Now, taking a0= 1 for y1, what is a2? Using that, what is a4?

Taking a1= 1 for y2, what is a3? Using that, what is a5.