Why using series solution

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Q1 :For example
Solve the D.E : U''-2xU' +2u = 0
Do I write out the series solution or write somthing like : C1e^r1 + c2e^r2 ,for r1&r2 are the roots of the equation~ What is the different between these two answer?


One more question,
Q2 : Show function (arcsin x )^2 satisfies
(1-x^2)U'' - xU' = 2 /For U(0) = 0; U'(0) = 0
Can write out few steps for me please?
 
yukcream said:
Q1 :For example
Solve the D.E : U''-2xU' +2u = 0
Do I write out the series solution or write somthing like : C1e^r1 + c2e^r2 ,for r1&r2 are the roots of the equation~ What is the different between these two answer?


One more question,
Q2 : Show function (arcsin x )^2 satisfies
(1-x^2)U'' - xU' = 2 /For U(0) = 0; U'(0) = 0
Can write out few steps for me please?
I'm not sure about using a series solution about the first (I totally forgot how to do those :() but make the substitution u = e^kx, and thus find u' and u'', substitute and solve the quadratic equation giving you the roots... and the solution would be something like Ae^r1x + Be^r2x where r1 and r2 are the roots. (But if they're equal roots, then it's Ae^r1x + Bxe^r1x I think)
I think the difference is just the method.... but hey I could be wrong.

Not sure about Q2 sorry. :(

EDIT: Oh wait, didn't see the x in the co-efficient...
Disregard everything I said! :P
 
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lurflurf

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yukcream said:
Q1 :For example
Solve the D.E : U''-2xU' +2u = 0
Do I write out the series solution or write somthing like : C1e^r1 + c2e^r2 ,for r1&r2 are the roots of the equation~ What is the different between these two answer?


One more question,
Q2 : Show function (arcsin x )^2 satisfies
(1-x^2)U'' - xU' = 2 /For U(0) = 0; U'(0) = 0
Can write out few steps for me please?
Q1
Assuming x is the independent variable and not a constant, the equation does not have constant coefficients. That is the c1e^r1 thing will not work. Series solution will work. Another choice given how simple the equation is, is to differentiate it, this gives an easy to solve third order equation. Though the solution will require an antiderivative of e^(x^2) which does not have an elementary antiderivative. You could make use of a special function such as erfi(x), or leave the integral in your answer. Also you will have one too many constants in the answer (3 instead of 2) eliminate the extraneous one by requiring the function to satisfy the orginal equation. Another route would be to see (by inspection) that U=x is a solution and use the method of reduction of order, an antiderivative of exp(x^2) is still required. The difference between the answers of the two methods is one is a nonelementary function and the other is an infinite series, but both represent the same thing.

Q2 simply find U' and U'' using the chain rule, power rule ect.
recall from elementary calculus that
Arcsin'(x)=1/sqrt(1-x^2) (for -1<x<1)
then check that the equation and conditions are meet.
 
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HallsofIvy

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yukcream said:
Q1 :For example
Solve the D.E : U''-2xU' +2u = 0
Do I write out the series solution or write somthing like : C1e^r1 + c2e^r2 ,for r1&r2 are the roots of the equation~ What is the different between these two answer?
I assume you mean something like
[tex]\Sigma_{n=0}^\infty a_nx^n[/tex]
Yes, you could do this as an infinite series but it would be one heckuvalot of work and the answer would be difficult to understand. Of course, the infinite series would be just a combination of the Taylor's series for er1x and er2x (don't forget the variable!).


One more question,
Q2 : Show function (arcsin x )^2 satisfies
(1-x^2)U'' - xU' = 2 /For U(0) = 0; U'(0) = 0
Can write out few steps for me please?
Suppose you were asked to show that x= 1 satisfies x5- 3x3+ 2x2+ x- 4= 0? You would just replace x with 1:
1- 3+ 3+ 2+ 1- 4= 0! That's a lot easier than solving the equation in the first place!

To show that (arcsin x)2 satisfies (1- x2)U"- xU'= 2,
Let U= (arcsin x)2, find the first and second derivatives of U and plug them into the equation. Of course, you would also have to show that this U satisfies U(0)= 0 (arcsin(0)= 0 so U(0)= (arcsin(0))2= 02= 0) and U'(0)= 0.
 

saltydog

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yukcream said:
Q1 :For example
Solve the D.E : U''-2xU' +2u = 0
Do I write out the series solution or write somthing like : C1e^r1 + c2e^r2 ,for r1&r2 are the roots of the equation~ What is the different between these two answer?


One more question,
Q2 : Show function (arcsin x )^2 satisfies
(1-x^2)U'' - xU' = 2 /For U(0) = 0; U'(0) = 0
Can write out few steps for me please?
By inspection y=x is a particular solution to the first. Then use reduction of order to find the general solution. Also series works well. I say work it both ways and verify that the series solution is equal to the one you get by reduction of order.


The second one, just differentiate and then back-substitute. You know:

[tex](1-x^2)\frac{d^2}{dx^2}ArcSin^2(x)-x\frac{d}{dx}ArcSin^2(x)[/tex]

Edit: Guess you guys already said that. I'm leaving mine cus' I think its a good idea to work it both ways and see how the answers compare. Which I'll do. :smile:
 
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HallsofIvy

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I completely overlooked those "x"s in the coefficients! Yes, lurflurf is right. "Series solutions" are used for linear equations (they don't work, generally, for non-linear equations) in which the coefficients are not constant (and so solving the "characteristic equation" does not work).

For the second equation, just write (1- x2)U" as U"- x2U" and proceed as in the first.
 

saltydog

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Via reduction of order:

[tex]y^{''}-2xy^{'}+2y=0[/tex]

Letting:

[tex]y=xv(x)[/tex]

The equation reduces to:

[tex]v^{''}+\frac{2(1-x^2)}{x}v^{'}=0[/tex]

Letting:

[tex]w=v^{'}[/tex]

give:

[tex]w^{'}+\frac{2(1-x^2)}{x}w=0[/tex]

The integrating factor is:

[tex]\sigma=x^2e^{-x^2}[/tex]

yielding:

[tex]d\left(x^2e^{-x^2}w\right)=0[/tex]

Integrating:

[tex]w=\frac{c_1e^{x^2}}{x^2}[/tex]

Since [itex]w=v^{'}[/itex] then:

[tex]dv=\frac{c_1e^{x^2}}{x^2}dx[/tex]

Integrating by parts leaves:

[tex]v=c_1\left(-\frac{e^{x^2}}{x}+2\int e^{x^2}dx\right)\quad \tag{1}[/tex]

Noting once again that:

[tex]Erfi[x]=\frac{2}{\sqrt{\pi}}\int_0^x e^{t^2}dt[/tex]

(1) can be written as:

[tex]v=c_1\left(\sqrt{\pi}Erfi[x]-\frac{e^{x^2}}{x}\right)+c_2[/tex]

Since y=xv we have finally the solution:

[tex]y(x)=c_1\left(x\sqrt{\pi}Erfi[x]-e^{x^2}\right)+c_2x[/tex]
 
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saltydog

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Alright, I think I have it:

Using power series, I obtain for the ODE:

[tex]a_0:\quad\text{arbitrary}[/tex]

[tex]a_1:\quad\text{arbitrary}[/tex]

[tex]a_2=-a_0[/tex]

[tex]n\geq3:\quad a_n=\frac{2a_{n-2}(n-3)}{n(n-1)}[/tex]

Now, it takes some doin' but the series solution can be put into the form:

[tex]y(x)=a_0+a_1x-\sum_{n=0}^{\infty}\frac{2^{n+1}a_0(1\cdot 3\cdot 5 \cdot 7\ldots(2n-1))x^{2n+2}}{(2n+2)!}[/tex]

This can be written as:

[tex]a_1x+a_0\left[1-x^2-\sum_{n=1}^\infty \frac{2^{n+1}(1\cdot 3\cdot 5\cdot 7\ldots(2n-1))x^{2n+2}}{(2n+2)!}\right][/tex]

or:

[tex]a_1x+a_0\left[1-x^2-\frac{x^4}{6}-\frac{x^6}{30}-\frac{x^8}{168}-\ldots\right]\tag{1}[/tex]

Now consider the solution from reduction of order:

[tex]y(x)=c_1\left[x\sqrt{\pi}Erfi[x]-e^{x^2}\right]+c_2x[/tex]

Converting Erfi[x] back to an integral results in:

[tex]y(x)=c_1\left[2x\int_0^x e^{t^2}dt-e^{x^2}\right]+c_2x[/tex]

Expanding the integrand into a power series:

[tex]y(x)=c_1\left[2x\int_0^x \left(1+t^2+\frac{t^4}{2}+\frac{t^6}{6}+\frac{t^8}{24}\ldots\right) dt-e^{x^2}\right]+c_2x[/tex]

Evaluating the integral and expanding [itex]e^{x^2}[/tex]:

[tex]y(x)=c_1\left[2x^2+\frac{2x^4}{3}+\frac{2x^6}{10}+\frac{2x^8}{42}+\ldots-
\left(1+x^2+\frac{x^4}{2}+\frac{x^6}{6}+\frac{x^8}{24}+\ldots\right)\right]+c_2x[/tex]

Simplifying common powers of x results in:

[tex]y(x)=c_1\left[-1+x^2+\frac{x^4}{6}+\frac{x^6}{30}+\ldots\right]+c_2x[/tex]

This is identical to (1) above.
 

lurflurf

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Saltydog you gave away the farm! Here is another simple method
U''-2xU'+2U=0
differentiate
U'''-2xU''=0
easy to solve
U=c1(I^2)exp(x^2)+c2x+c3
where I means take an anti derivative and I^2 means take two antiderivatives
clearly c3=0 and is an extraneous solution
U=c1(I^2)exp(x^2)+c2x
integrating and changing arbitrary constant
U=c1 I erfi(x)+c2x
integrating by parts and changing arbitrary constant
U=c1(sqrt(pi)x erfi(x)-exp(x^2))+c2x
The same as other methods give
where erfi(x) is a know special function whose derivative is a constant multiple of exp(x^2)
 

saltydog

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lurflurf said:
Saltydog you gave away the farm!
Well I hope not. :uhh: Yukcream, you know how to do all the stuf up there I did? Ask questions if not. Hey, I don't have a clue how to do that one over there dealing with functionals!

Can you condense the series solution into that summation? Usually, figure out 10 or so of them by hand. In the case above, the odds ones drop out. Then look at them and start fooling around with the numbers and try to see a trend. I bet if you can't quite get it you can get it close, at least the powers on x, the denominator, powers on 2 maybe. The 1,3,5,7 and so on is a little more tricky. And if I did give it away, well I say do another one like:

[tex]y^{''}+2xy^{'}-2y=0[/tex]

:smile:
 

lurflurf

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saltydog said:
And if I did give it away, well I say do another one like:
[tex]y^{''}+2xy^{'}-2y=0[/tex]
That is not much different, but could be fun.
Problems of this type border on the interesting area of boundry layer analysis.
consider
hu''+xu'-u=0 u(0)=u(1)=1
for h a very small positive number
the solutions for x not small is much like the solution of
xu'-u=0 u(1)=1
but at the last second the function goes oh yeah I forgot about the other boundry condition and swoops to it.
Another given in Haberman's EAPDE is
hu''-u'+2xu=0 u(0)=3 u(1)=2
for h a very small positive number
which except very near 1 is much like
-u'+2xu=0 u(0)=3
but it swoops to u(1)=2 quickly
Habermans example involves hideous Airy functions
 

saltydog

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lurflurf said:
Another given in Haberman's EAPDE is
hu''-u'+2xu=0 u(0)=3 u(1)=2
for h a very small positive number
which except very near 1 is much like
-u'+2xu=0 u(0)=3
but it swoops to u(1)=2 quickly
Habermans example involves hideous Airy functions
This one was more of a challenge:

[tex]y^{''}-xy=0[/tex]

Via power series, I got:

[tex]
\begin{align*}
y(x)=a_0&+\sum_{k=1}^{\infty}\frac{a_0(1\cdot 4\cdot 7\cdot 10\cdot\cdot\cdot (3k-2))x^{3k}}{(3k)!} \\
&+a_1x+\sum_{k=1}^{\infty}\frac{a_1(2\cdot 5\cdot 8\cdot 11\cdot\cdot\cdot (3k-1))x^{3k+1}}
{(3k+1)!}
\end{align}
[/tex]

or just let a0 and a1 be 1 and then let:

[tex]y(x)=c_1f_1(x)+c_2f_2(x)[/tex]

with [itex]f_1(x)[/itex] and [itex]f_2(x)[/itex] equal to the [itex]a_0[/itex] and [itex]a_1[/itex] expressions respectively.

Now,the solution can also be expressed in terms of Airy functions:

[tex]y(x)=c_1Ai(x)+c_2Bi(x)[/tex]

One reference I have for Ai(x) is:

[tex]Ai[x]=\frac{1}{\sqrt{\pi}}\int_0^{\infty} Cos\left[\frac{v^3}{3}+vx\right]dv[/tex]

So I suppose one of those functions up there could be converted to that one although I don't know how.
 

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