First Order Seires Solution ODE

[EnzoF61]

Homework Statement

y' = $$\sqrt{(1-y^2) }$$
Initial condition y(0) = 0
a) Show y = sinx is a solution of the initial value problem.
b) Look for a solution of the initial value problem in the form of a power series about x = 0. Find coefficients up to the term in x^3 in this series.

Homework Equations

part a) was Ok.

The Attempt at a Solution

This is for my part b attempt.

(y')^2 + (y)^2 = 1

(cosx)^2 + (sinx)^2 = 1

$$\sum(-1)^ {2n} * (x^ {4n} )}/(2n!)^2$$ + $$\sum (-1)^{2n} * (x^ {4n+2})/((2n+1)!)^2$$ = 1

I had also tried the general solution series for y and y'. y = a0 + a1x + ... + an*x^n
y' = a1 + ... + n*an x^n
Please note the values such as a0, a1, ...an have the nth term as a subscript.

Maybe I should not be using part a where y=sinx and y'=cosx because I will have even powers when squared always...?

What series will allow me to evaluate at the x^3 term?

 

[Mark44]

In your work for part b, you are assuming that you already have a solution. Start with y(x) = a_0 + a_1x +a_2x^2 + ... + a_nx^n + ...

You don't need very many terms in the series above, since you need to work only up to the 3rd degree term in your solution.

 

[EnzoF61]

Hmm

Let y = sinx be a solution of y' = sqrt(1-y^2)

integral ( y') = integral {sqrt( 1 - sin^2(x) )}

integral (dy) = intergral {cos(x) dx}

y = sin(x)

Given y(0) = 0

We have that,
y(0) = sin(0) = 0

==> y = sin(x) is a solution of the initial value problem

Hence y = summation ( (-1)^n * x^ 2n+1 )/ (2n+1)!
for the terms to x^3 we have that summation ( (-1)^n * x^ 2n+1 )/ (2n+1)! = x - x^3/6

I know you're not my instructor would I be able to say from part a) we know that y = sinx is a solution Hence y = summation ( (-1)^n * x^ 2n+1 )/ (2n+1)!
for the terms to x^3 we have that summation ( (-1)^n * x^ 2n+1 )/ (2n+1)! = x - x^3/6

 

[Mark44]

Hmm

Let y = sinx be a solution of y' = sqrt(1-y^2)

NO!
In your first post it says
"b) Look for a solution of the initial value problem in the form of a power series about x = 0."

I already told you in my previous post what you need to do to accomplish that.

integral ( y') = integral {sqrt( 1 - sin^2(x) )}

integral (dy) = intergral {cos(x) dx}

y = sin(x)

Given y(0) = 0

We have that,
y(0) = sin(0) = 0

==> y = sin(x) is a solution of the initial value problem

Hence y = summation ( (-1)^n * x^ 2n+1 )/ (2n+1)!
for the terms to x^3 we have that summation ( (-1)^n * x^ 2n+1 )/ (2n+1)! = x - x^3/6

I know you're not my instructor would I be able to say from part a) we know that y = sinx is a solution Hence y = summation ( (-1)^n * x^ 2n+1 )/ (2n+1)!
for the terms to x^3 we have that summation ( (-1)^n * x^ 2n+1 )/ (2n+1)! = x - x^3/6

 

[EnzoF61]

y = n=0--> ∞ sum(a_n*x^n)

y' = n=1-->∞ sum(n*a_n*x^n-1)

y' = sqrt(1-y^2)

y = sqrt(1-(y')^2)

y(0) = 0

sqrt(1-(y')^2) = 0

(n=1 --> ∞ sum(n*a_n*x^n-1))^2 = 1

Solved for y = 0

Now I have sum for (y')^2 = 1

I am having ambiguity with the general series for y(x) = a_0 + a_1x +a_2x^2 + ... + a_nx^n + ...

 

[Mark44]

1. Start with y(x) = a_0 + a_1x +a_2x^2. That should probably be enough terms.
2. Calculate (y(x))^2
3. Calculate 1 - (y(x))^2
4. Calculate y'(x)
5. Calculate (y'(x))^2

Equate the expressions you got in the last two steps and find the coefficients up to the 3rd degree term. What you're doing in the last two steps is setting y'^2 = 1 - y^2, which seems easier than setting y' = sqrt(1 - y^2). The two are not equivalent, so it's possible you'll get a solution of y'^2 = 1 - y^2 that is not also a solution of your original differential equation.

 

[EnzoF61]

I believe that it is impossible to obtain a recursion series solution up to the x^3 term.

However I did recognize that

y' = sqrt(1-y^2)

dy/dx = sqrt(1-y^2)

dy/sqrt(1-y^2) = dx

integral (dy/sqrt(1-y^2)) = integral (dx)

inverse sin (y) = x

y = sin(x)

y(0) = 0

y = sin(0) = 0

Hence the series solution is y = sinx = series(sinx) = x - (x^3)/6 for values up to the x^3.

 

[Mark44]

In my opinion, that is NOT what you are supposed to do for part b. You are supposed to start with a power series, not find a solution and then write that as a power series. I have told you what I believe you need to do in posts #2, #4, and #6.

 

[EnzoF61]

I don't think that it is possible start with a series solution. The question says look for a solution of the initial value problem in the form of a power series. Not only can we can show that this is a solution of the initial value problem, but by using Taylor's expansion, sin(x) is in the form of a well known power series. Expanding the terms for a general summation of y an y' will not lead to a recursion formula, which is necessary to find a power series. This is sure to be the only way to find a solution.

 

[Mark44]

I don't think that it is possible start with a series solution.

Apparently you don't.

The question says look for a solution of the initial value problem in the form of a power series.

YES! And that means that you should START with a GENERAL power series. The differential equation y' = sqrt(1 - y²) is very simple to solve by separation of variables, but I don't believe that is the intent of part b of this problem.

Not only can we can show that this is a solution of the initial value problem, but by using Taylor's expansion, sin(x) is in the form of a well known power series. Expanding the terms for a general summation of y an y' will not lead to a recursion formula, which is necessary to find a power series.

Not so. You don't need a recursion formula - all you need are the terms up to degree three.

This is sure to be the only way to find a solution.

I disagree - I have shown you a different approach, but have it your way, I'm done here. You asked for help, and I provided some help in the direction I believe this problem wants you to go. If you choose not to take the advice I offered, that's your choice.

 

[EnzoF61]

I have still been working with this problem. I've tried general series, squaring them, binomial expansions. Equating coefficients. This is IMPOSSIBLE to solve numerically.

 

[EnzoF61]

The homework has already been turned in, but I challenge you to prove me wrong and solve this numerically.

http://www.math.pitt.edu/~troy/classes/math1270.html

You can see that Homework 3 was due on the 18th
The book was Boyce and DiPrima 9th Ed. Elementary Diff Eq and Boundary Value Problems.

 

[Mark44]

I have still been working with this problem. I've tried general series, squaring them, binomial expansions. Equating coefficients. This is IMPOSSIBLE to solve numerically.

It's impossible the way you've been trying to do it, which is to approach it with a general series. It's not impossible if you take only a small number of terms in your series, say three at the most.

 

[EnzoF61]

I'm coming closer, but my results are still not precise.

y = a_0 + a_1x + a_2x^2

y(0) = 0

y(0) = a_0 + 0 + 0 = 0 ==> a_0 = 0

y = a_1x + a_2x^2
y' = a_1 + 2a_2x

(y)^2 = (a_1)^2*x^2 + 2a_1*a_2*x^3 + (a_2)^2*x^4
(y')^2 = (a_1 + 2a_2x)^2


(y')^2 = 1 - (y)^2

(a_1 + 2a_2x)^2 = 1 - {(a_1)^2*x^2 + 2a_1*a_2*x^3 + (a_2)^2*x^4}

y' = 1 - {(a_1)^2*x^2 + 2a_1*a_2*x^3 + (a_2)^2*x^4} / {(a_1 + 2a_2x)}

dy/dx = 1 - {(a_1)^2*x^2 + 2a_1*a_2*x^3 + (a_2)^2*x^4} / {(a_1 + 2a_2x)}


INT [(a_1 + 2a_2x) dy] = INT[1 - {(a_1)^2*x^2 + 2a_1*a_2*x^3 + (a_2)^2*x^4}]

a_1y + 2a_2xy = x - [(a_1)^2*x^3] /3

y(a_1 + 2a_2x) = x - [(a_1)^2*x^3] /3

I see it, but I'm stuck at this moment. Please help if I have a particular step incorrect.

 

[Mark44]

I'm coming closer, but my results are still not precise.

y = a_0 + a_1x + a_2x^2

y(0) = 0

y(0) = a_0 + 0 + 0 = 0 ==> a_0 = 0

y = a_1x + a_2x^2
y' = a_1 + 2a_2x

(y)^2 = (a_1)^2*x^2 + 2a_1*a_2*x^3 + (a_2)^2*x^4
(y')^2 = (a_1 + 2a_2x)^2


(y')^2 = 1 - (y)^2

(a_1 + 2a_2x)^2 = 1 - {(a_1)^2*x^2 + 2a_1*a_2*x^3 + (a_2)^2*x^4}

Looks fine up to here. Now move everything over to one side so that 0 is on the other side, and group together terms in the same power of x. What you'll have on the left side is a polynomial in x, and on the right side, another polynomial in x, all of whose coefficients are zero.

Now equate the coefficients of 1, x, x^, and x^3 on both sides to solve for your a_1, a_2, and a_3. They will give you your coefficients for you Maclaurin series, which is the goal of this whole exercise.

y' = 1 - {(a_1)^2*x^2 + 2a_1*a_2*x^3 + (a_2)^2*x^4} / {(a_1 + 2a_2x)}

dy/dx = 1 - {(a_1)^2*x^2 + 2a_1*a_2*x^3 + (a_2)^2*x^4} / {(a_1 + 2a_2x)}


INT [(a_1 + 2a_2x) dy] = INT[1 - {(a_1)^2*x^2 + 2a_1*a_2*x^3 + (a_2)^2*x^4}]

a_1y + 2a_2xy = x - [(a_1)^2*x^3] /3

y(a_1 + 2a_2x) = x - [(a_1)^2*x^3] /3

I see it, but I'm stuck at this moment. Please help if I have a particular step incorrect.

 

[EnzoF61]

Now equate the coefficients of 1, x, x^, and x^3 on both sides to solve for your a_1, a_2, and a_3. They will give you your coefficients for you Maclaurin series, which is the goal of this whole exercise.

Are you interpreting a_3 as another coefficient in my work? I.e. a_1*a_2 = a_3

 

[Mark44]

No, I meant a_3 as the coefficient of x^3 in the series.