Compute the solution of y"' - xy' = 0

[bballing1210]

Hi all,

I've been having trouble answering these two ODE problems. Hopefully someone can help me out.

1. Compute the solution of y"' - xy' = 0 which satisfies y(0) = 1, y'(0) = 0, and y"(0) = 0.

I've tried using the power series expansion for y and substituting it in and getting the recurrence relation, but when I substitute back the initial conditions I'm getting the two series cancelling out which I don't think is right.

2. Solve the initial value problem 3y" - y' + (x+1)y = 1 with y(0) = y'(0) = 0

For this one, I know you have to compute both the power series expansion as well as a particular solution through substitution to be able to apply the initial conditions, but I'm seriously stuck in even thinking about what to substitute, let alone how to tackle this. I tried to start developing out the recurrence relationship, but it got really messy and I don't think I'm doing it right.

Thanks a lot to whoever can help, I really appreciate it!

 

[HallsofIvy]

Hi all,

I've been having trouble answering these two ODE problems. Hopefully someone can help me out.

1. Compute the solution of y"' - xy' = 0 which satisfies y(0) = 1, y'(0) = 0, and y"(0) = 0.

I've tried using the power series expansion for y and substituting it in and getting the recurrence relation, but when I substitute back the initial conditions I'm getting the two series cancelling out which I don't think is right.

Yes, that's exactly what has to happen! The first thing I would do is let u= y' so the equation becomes u''= xu, u(0)= 0, u'(0)= 0. That satisfies all the conditions for "existence and uniqueness" of solutions and u(x)= 0 for all x is an obvious solution. That means that y is the constant function: y(x)= 1 for all x satisfies the differential equation and initial conditions.

2. Solve the initial value problem 3y" - y' + (x+1)y = 1 with y(0) = y'(0) = 0

For this one, I know you have to compute both the power series expansion as well as a particular solution through substitution to be able to apply the initial conditions, but I'm seriously stuck in even thinking about what to substitute, let alone how to tackle this. I tried to start developing out the recurrence relationship, but it got really messy and I don't think I'm doing it right.

Thanks a lot to whoever can help, I really appreciate it!

Doesn't look too bad. Letting $y= \sum_{n=0}^\infty a_nx^n$, $y'= \sum_{n=1}^\infty na_nx^n$ and $y''= \sum_{n=2}^\infty n(n-1)a_nx^{n-2}$ so the equation becomes
$$3\sum_{n=2}^\infty n(n-1)a_nx^{n-2}- \sum_{n=1}^\infty na_mx^{n-1}+ \sum_{n=0}^\infty a_mx^{n+1}+ \sum_{n=0}^\infty a_nx^n= 1$$
To get the same powers, change the dummy indices: in the first sum, let j= n-2 so that n= j+ 2 and it becomes
$$3\sum_{j=0}^\infty (j+2)(j+1)a_{j+2}x^j$$
In the second sum, let j= n- 1 so it becomes
$$\sum_{j= 0}^\infty (j+1)a_{j+1}x^j$$
In the third sum, let j= n+ 1 so that n= j- 1 and it becomes
$$\sum_{j= 1}^\infty a_{j-1}x^j$$.
Finally, in the fourth sum, let j= n so it becomes
$$\sum_{j= 0}^\infty a_jx^j$$

Note that the third sum does not start until j= 1 (since it starts with $x^1$) so for j= 0 we have
$$6a_2- a_1+ a_0= 1$$
so that $a_2= (a_1- a_0+ 1)/6$
since we are given that $a_0= y(0)= 0$ and $a_1= y'(0)= 0$
that tells us that $a_2= 1/6$.

For j> 0, we have
$$3(j+2)(j+1)a_{j+2}- (j+1)a_{j+1}+ a_{j-1}+ a_j= 0$$
or
$$a_{j+2}= \frac{(j+1)a_{j+1}- a_{j-1}- a_j}{3(j+2)(j+ 1)}$$

Then $a_3= (2a_2- a_0- a_1)/18)= 1/18$, $a_4= (2a_3- a_1- a_2)/36= -1/648$, $a_5= (2a_4- a_2- a_3)/90$ etc. Yes, it wil probably be difficult to find a general formula. Are you required to?

 

[bballing1210]

Hi. Thanks a lot for the help!

For number 2, what happened to the coefficient of 3 in front of the y"? You don't factor this into the series solution?

And I think we're allowed to keep the answers in a recurrence relation as long as it's correct, so I think that's what I'll do and I'll list out some terms in the series for clarification. Thanks!

 

[HallsofIvy]

Yes, I accidently dropped the 3. I've gone back and corrected that. Hope its right this time!

 

[blue_raver22]

I would suggest using the method of Frobenius to solve these ODEs. This method involves finding a series solution for y and then substituting it back into the original equation to determine the recurrence relation and solve for the coefficients.

For the first problem, we can start by assuming a series solution of the form y = ∑anxn. Substituting this into the equation, we get:

∑n(n-2)(n-1)anxn-2 - x∑nanxn = 0

We can then simplify this to:

∑n(n-2)(n-1)anxn-2 - ∑n(n-1)anxn = 0

Next, we can separate out the first few terms and use the initial conditions to determine the coefficients:

a0 = 1
a1 = 0
a2 = 0

Substituting these values back into the equation, we get:

2a3 - a2 = 0
a3 = 0

Continuing this process, we can see that all coefficients after a2 will be zero, resulting in y = 1 as the solution.

For the second problem, we can use a similar approach. Assuming a series solution of the form y = ∑anxn, we can substitute it into the equation and simplify to get:

∑n(n-1)(3an+1)xn-1 + ∑nanxn+1 = 1

Using the initial conditions, we can determine the first few coefficients:

a0 = 0
a1 = 0
a2 = 1/6

Substituting these values back into the equation, we can solve for the remaining coefficients:

a3 = -1/120
a4 = 1/720

Thus, the solution is y = 1/6x^2 - 1/120x^3 + 1/720x^4 + ... which satisfies the initial conditions.

In conclusion, using the method of Frobenius can help us solve these ODEs and determine the series solutions that satisfy the given initial conditions. It may seem daunting at first, but with practice, it becomes easier to apply.