Series Solutions

Differential Equations Project

Objectives:

(a)   To understand the use of power series in approximating a solution to a differential equation

       whose coefficients are functions of the independent variable.

(b)   To introduce the series  option of the   dsolve   command, as well as the commands Order

        and convert .

     

            The power series method will allow us to handle differential equations with variable

       coefficients, most of which cannot be solved in closed form; but this method  involves quite a

       bit of computation.  Maple has built-in   capabilities that ease the computational burden and

       allow us to focus on the meaning of the approximate solution.

Solved Example 1:

Solve the initial value problem   y ''  -   x   y '  +  4 y  ;    y (0) = 1,    y '(0) = 0.

Solution:

First we note that the equation does not have constant coefficients.  Next we see that x  = 0  is an ordinary point  of the equation.  Existence and uniqueness theory guarantees that this IVP has a unique solution.  The theory of series solutions also guarantees that the solution has a Maclaurin series expansion that represents the solution at every value of   x .  To control the size of the degree of the polynomial approximation, we invoke the Order  command, first asking to see all the terms of the series up to order 3 and then up to order 10.  We use the series  option in dsolve :

>    with(DEtools):

>    Order:=3:
de:=diff(y(x),x$2)-x*diff(y(x),x)+4*y(x);

de := diff(y(x),`$`(x,2))-x*diff(y(x),x)+4*y(x)

>    dsolve({de,y(0)=1,D(y)(0)=0}, y(x), type=series);

y(x) = series(1-2*x^2+O(x^3),x,3)

>    Order:=10:

>    dsolve({de,y(0)=1,D(y)(0)=0}, y(x), type=series);

y(x) = 1-2*x^2+1/3*x^4

>   

>    Order:=20:

>    dsolve({de,y(0)=1,D(y)(0)=0}, y(x), type=series);

y(x) = 1-2*x^2+1/3*x^4

It seems as if the Maclaurin series for the solution is just the polynomial   1-2*x^2+x^4/3  .  We can check this and see that we have been lucky enough to find a simple closed form solution in the form of a fourth degree polynomial.

Solved Example 2:

Solve the equation   (5*x^2-2)*diff(y,`$`(x,2))+15*x*diff(y,x)+5*y = 0   near the point   x  = 0.

Solution:

The only singular points are   x = sqrt(2/5)   and   x = -sqrt(2/5)  , so that   x  = 0 is an ordinary point.  Therefore we can represent the general solution of our equation as a Maclaurin series with radius of convergence at least   sqrt(2/5)   .  Let's get an idea of the form of the series solution:

>    Order:=10:

>    de2:=(5*x^2-2)*diff(y(x),x$2)+15*x*diff(y(x),x)+5*y(x)=0;

de2 := (5*x^2-2)*diff(y(x),`$`(x,2))+15*x*diff(y(x),x)+5*y(x) = 0

>    dsolve(de2,y(x),type=series);

y(x) = series(y(0)+D(y)(0)*x+5/4*y(0)*x^2+5/3*D(y)(0)*x^3+75/32*y(0)*x^4+10/3*D(y)(0)*x^5+625/128*y(0)*x^6+50/7*D(y)(0)*x^7+21875/2048*y(0)*x^8+1000/63*D(y)(0)*x^9+O(x^10),x,10)
y(x) = series(y(0)+D(y)(0)*x+5/4*y(0)*x^2+5/3*D(y)(0)*x^3+75/32*y(0)*x^4+10/3*D(y)(0)*x^5+625/128*y(0)*x^6+50/7*D(y)(0)*x^7+21875/2048*y(0)*x^8+1000/63*D(y)(0)*x^9+O(x^10),x,10)

>   

>    convert(rhs(%), polynom);

y(0)+x*D(y)(0)+5/4*y(0)*x^2+5/3*D(y)(0)*x^3+75/32*y(0)*x^4+10/3*D(y)(0)*x^5+625/128*y(0)*x^6+50/7*D(y)(0)*x^7+21875/2048*y(0)*x^8+1000/63*D(y)(0)*x^9

  

Solved Example 3:

Use a series method to find the general solution of   2*x*diff(y,`$`(x,2))+(1-2*x^2)*diff(y,x)-4*x*y = 0 ,   x  > 0, near the point   x  = 0.

Solution :

First we note that   x  = 0  is a regular   singular point .  Although a regular power series will not work, we could try to find solutions in the form of a Frobenius series    y(x) = sum(c[n]*x^(n+r),n = 0 .. infinity) , where  the number   r   satifies a certain   indicial equation.  Maple handles all this with ease:

>    with(DEtools):

>    DE:=2*x*diff(y(x),x$2)+(1-2*x^2)*diff(y(x),x)-4*x*y(x)=0;

DE := 2*x*diff(y(x),`$`(x,2))+(1-2*x^2)*diff(y(x),x)-4*x*y(x) = 0

>    dsolve(DE,y(x),type=series);

y(x) = _C1*x^(1/2)*(series(1+1/2*x^2+1/8*x^4+1/48*x^6+1/384*x^8+O(x^10),x,10))+_C2*(series(1+2/3*x^2+4/21*x^4+8/231*x^6+16/3465*x^8+O(x^10),x,10))

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ASSIGNMENT

Problem 1:

Find the power series solution of  the initial value problem

                             y ''  -   x y '  -   y   =  0 ;   y (0) = 2,   y '(0) = 1,

  

Problem 2:

Find the general solution near   x  = 0  of    (4-x^2)*diff(y,`$`(x,2))-2*x*diff(y,x)+2*y = 0 .

Problem 3:

Consider the equation    2*x^2*(1-x)*diff(y,`$`(x,2))-x*(1+x)*diff(y,x)+(1+x)*y = 0 ,   x  > 0.

(a)   Is   x  = 0  a regular point or a singular point of the equation?

 

(b)   Find the general solution of the equation in terms of a series centered at   x  = 0.

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   MSIP Grant #P120A80089-98: "Three Urban Calculus Reform Programs: Adopting the Best,"    1998-2001, Revised September 18, 2003