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Series Solution of a DE

weber

New member
Jun 24, 2012
3
Hey!

I'm having problems with finding the general solution of this DE,

using series.

I have readed the Zill book, but I don't know how to start solving.

Any help is appreciated!

[TEX] y'' - 4xy' -4y = e^x[/TEX]
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hey!

I'm having problems with finding the general solution of this DE,

using series.

I have readed the Zill book, but I don't know how to start solving.

Any help is appreciated!

[TEX] y'' - 4xy' -4y = e^x[/TEX]
Hi weber, :)

Let me provide some insight into this problem.

This is a Linear Non-homogeneous differential equation with variable coefficients. Although there are standard methods(which are mentioned in Zill's book) to solve Non-homogeneous differential equations with constant coefficients, this one does not fall into that category.

However you can proceed with the power series method by taking, \(y=\sum_{x=0}^{\infty}a_{n}x^n\) and \(e^x=\sum_{x=0}^{\infty}\frac{x^n}{n!}\)

I got the recurrence relation,

\[a_2=\frac{4a_0+1}{2}\mbox{ and }a_{n+2}=\frac{4(n+1)a_n+\frac{1}{n!}}{(n+1)(n+2)} \mbox{ for }n\geq 1\]

By looking at the solution that Wolfram gives, even if we can get a closed form for this recurrence relation it would need some tedious bit of algebra.

By clicking on the "show steps" button in the above link you may see that a much simpler method is suggested by Wolfram.

Kind Regards,
Sudharaka.
 

weber

New member
Jun 24, 2012
3
This helped me a lot!

Thank you again Sudharaka!
 

chisigma

Well-known member
Feb 13, 2012
1,704
Hey!

I'm having problems with finding the general solution of this DE,

using series.

I have readed the Zill book, but I don't know how to start solving.

Any help is appreciated!

[TEX] y'' - 4xy' -4y = e^x[/TEX]
It is requested the general solution of the second order DE...

$\displaystyle y^{\ ''} -4\ x\ y^{\ '} -4\ y=e^{x}$ (1)

... so that this 'solution' must contain two arbitrary constants. A relatively 'easy' although not very 'popular' way to met this goal is first suppose that the general solution is analytic in x=0, so that is...

$\displaystyle y(x)=a_{0}+ a_{1}\ x + a_{2}\ x^{2}+ a_{3}\ x^{3} + ...$ (2)

... where...

$\displaystyle a_{n}= \frac{1}{n!}\ \frac{d^{n}}{d x^{n}} f(x)_{x=0}$ (3)

The 'arbitrary constants' are given by the initial conditions so that is $\displaystyle a_{0}=y(0)$ and $\displaystyle a_{1}=y^{\ '}(0)$. The other $a_{n}$ are derived from (1) as follows...

$\displaystyle y^{\ ''} = 4\ x\ y^{\ '} + 4\ y + e^{x} \implies a_{2}= \frac{1}{2}\ (4\ a_{0}+1)$ (4)

$\displaystyle y^{\ '''} = 4\ x\ y^{\ ''} +8\ y^{\ '} + e^{x} \implies a_{3}= \frac{1}{6}\ (8\ a_{1} + 1)$ (5)

$\displaystyle y^{(4)} = 4\ x\ y^{\ '''} +12\ y^{\ ''} + e^{x} \implies a_{4}= \frac{1}{24}\ (48\ a_{0}+13)$ (6)

... and the procedure can be repeated indefinitely...

Kind regards

$\chi$ $\sigma$
 
Last edited:

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
Just an added note. The ODE can be integrated once giving

$y' - 4xy = e^x + c_1$

which is now first order and linear.
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Another approach to "series solution" is to recall that we can always write a function, f(x), as its MacLaurin series $\sum_{i=0}^\infty \frac{f^{(n)}(0)}{n!} x^n$. If we are given the initial values as $y(0)= y_0$ and $y'(0)= y_1$, we can imediately use the differential equation to find y''(0): $y''= 4xy'+ 4y+ e^x$ so $y''(0)= 0(y_1)+ 4(y_0)+ 1= 4y_0+ 1$. Now, differentiate: $y'''= 4y'+ 4xy''+ 4y'+ e^x$ and taking x= 0, $y'''(0)= 8x_1+ 1$. Differentiate again: $y''''= 12y''+ 4xy'''+ e^x$ so that y''''(0)= 12(4_0+ 1)+ 1. Continue like that.