Find General Solution to Ordinary Differential Equation Calculator

Series Solutions to Differential Equations.

Prof. C. Madigan

Nova Scotia Agricultural College

Truro, N.S. B2N 5E3

cmadigan@nsac.ca

Solving linear differential equations with constant coefficients reduces to an algebraic problem. There is no similar procedure for solving linear differential equations with variable coefficients. With the exception of special types, such as the Cauchy equations, these will generally require the use of the power series techniques for a solution.

Initialization.

Review of Series and Power Series.

SERIES

Recall a power series in [powers of] x - a is an infinite series of the form

If a = 0, this is a power series in x

An important aspect of any series is whether or not it converges (IE the infinite sum exists). (one might say that the only good series is a converging series)

NOTE : A converging series can then be approximated by using its nth partial sum.

Definition 1 . Given an infinite seriesLet the sequence of partial sums of the series be . If this sequence is convergent IE if the series is called convergent and S is called its sum. If the limit does not exist, the series is called divergent .

Examples . 1. Geometric serieswhere a and r are constants. This series converges to if and diverges for The sum of the first n terms

2. p serieswhere p is a constant, converges for ", mathvariant = "normal", fence = "false", separator = "false", stretchy = "false", symmetric = "fal..." align="center" border="0"> and diverges for

Tests for convergence of a series

There are several tests for convergence of a series.

1. Comparison test for series of non-negative terms

Let and suppose converges. Then if 0≤ then converges.

If diverges and if then diverges.

2. Quotient test for series of non-negative terms.

Ifandand if either both converge or both diverge.

If A = 0 and if converges then converges

If A = ∞ and diverges then diverges

3. Integral test. for series of non-negative terms.

If f(x) is positive, continuous and monotonic decreasing for x ≥ N and is such that f(n) = , n = N, N+1, N+2, ...then converges or diverges according as converges or diverges.

4. Alternating Series test.

An alternating series converges if it satisfies the two conditions

a) for n ≥ 1

b) or

Definition: The series is called absolutely convergent if converges. If converges but diverges, then is called conditionally convergent.

NOTE: An absolutely convergent series is convergent.

5. Ratio test .

Let Then the series

a) converges (absolutely) if L < 1

b) diverges if L > 1

If L = 1 the test fails.

6. Raabe's test.

Let Then the series

a) converges (absolutely) if L > 1

b) diverges or converges conditionally if L < 1

If L = 1 the test fails.

7. Gauss' test.

Ifwhere for all n > N, then the series

a) converges absolutely if L > 1

b) diverges or is conditionally convergent if L ≤ 1

POWER SERIES

A power series converges at the point x = if the infinite series (of real numbers) converges.

In general a power series converges absolutely for and diverges for . R is called the radius of convergence.

Note: convergence at the endpoints requires separate analysis.

Example 1. Determine the convergence of

Applying the ratio test we have=

Hence the series converges absolutely for diverges for

For IE at we must test separately

At our series is the harmonic series known to diverge.

At x = $Typesetting:-mrow(Typesetting:-mfrac(Typesetting:-mn($ our series is an alternating harmonic series that converges.

The given series converges for $Typesetting:-mrow(Typesetting:-mfrac(Typesetting:-mn($ and diverges everywhere else.

Definition A function is said to be analytic at a point if, in an open interval about this function can be represented as a power series that has a positive radius of convergence.

Taylor Series If f is analytic at , then the representation

holds in some open interval about and is called the Taylor series for about When , it is also called the Maclaurin series for

Power Series Solutions to Linear Differential Equations.

We now consider a method for obtaining a power series solution to a linear differential equation with polynomial coefficients.

Given the differential equation

we begin by writing it in the standard form

and

Definition 1 A point is called an ordinary point of equation (1) if both p(x) and q(x) are analytic at If it is not an ordinary point, it is called a singular point of the equation.

Definition 2. A singular point of (2) is said to be a regular singular point if both and are analytic at Otherwise is called an irregular singular point.

Solutions about an ordinary point x =

We assume that a power series solution of the formexists and our task is to determine the coefficients This task is accomplished by substituting this series into the differential equation, combining the result into a single series by collecting the result in powers of x and then in order for this series to be identically zero, we must have that all of its coefficients must be equal to zero.

We make use of Maples commands , series, series ,coeff, coeff ,convert, convert collect, collect , etc to carry out the required work.

Example 1 Solve the equation 2y'' +xy' + y = 0 about the ordinary point x = 0.

Step 1 . Define the deq

(3.1.1.1)

Step 2. Define the series solution (begin by setting the order to be used for the series)

(3.1.1.2)

Step 3. Substitute the series into the deq.

Step 4 convert the series into a polynomial and collect in terms of powers of x

Step 5. Use the Maple command coeff to select the coefficients of the powers of x. An example of this command is

(3.1.1.6)

We then set the coefficients equal to zero and solve for the unknown derivatives i terms of the initial conditions that are required in order to solve a second order DEQ IE y(0) = a and y'(0) = b

Step 6. Finally we substitute these values into our solution and collect the terms in terms of the parameters a and b.

(3.1.1.8)

We can check our result using Maple's dsolve command

(3.1.1.9)

Example 2. Solve the differential equationabout the ordinary point x = 0.

Step 1 . Define the deq

(3.1.2.1)

Step 2 . Define the series solution (begin by setting the order to be used for the series)

Step 3 . Substitute the series into the equation

Step 4 convert the series into a polynomial and collect in terms of powers of x

Step 5 . Use the Maple command coeff to select the coefficients of the powers of x. Then set the coefficients equal to zero and solve for the unknown derivatives i terms of the initial conditions that are required in order to solve a second order DEQ ie y(0) = a and y'(0) = b

Step 6. Finally we substitute these values into our solution and collect the terms in terms of the parameters a and b.

We can check our result using Maple's dsolve command

NOTE: One of the difficulties is determining the radius of convergence when we are not able to find a general form for the coefficients and use the ratio test.

EXISTENCE OF ANALYTIC SOLUTIONS.

There is a theorem that states that if is an ordinary point then equation has two linearly independent analytic solutions of the form and that the radius of convergence is as least as large as the distance from to the nearest singular point (real or complex-valued) of the deq.

In example 1 above since there are no singular points the radius of convergence is ∞.

In example 2 there are two singular points � i and hence the radius of convergence of our solutions about the point x = 0 is at least 1.

Maple also has a power series package PowerSeries that allows you to work directly with power series.

We will use this package in the following examples.

Example 1. Solve the IVP y'' - (x-2)y' + 2y = 0; y(0) = 1, y'(0) = -1

We will use power series package in Maple to find the solution

First to create the series solution Ys(x) =

The command tpsform converts the Powseries created above into a power series form of the variable stated in the command.

(3.1.3.1)

Next we determine the necessary derivatives found in the given deq.

(3.1.3.2)

(3.1.3.3)

Note the answer is in the form of a procedure. You can see the power series by using the tpsform command.We also need to make the coeff of our eqn into a power series in order to use the powseries commands.

You can see the power series by using the tpsform command. For example

(3.1.3.4)

Next we combine these series to form the lhs of the equation given using the commands for multiplying and adding power series.

(3.1.3.5)

Using the op command to identify the coefficients , setting them equal to zero, and solving for them in terms of a[0] and a[1]

Substituting the coeff into Y(x)

Converting our solution into a polynomial and collecting in terms of

Finally, substituting the initial conditions and solving for a[0] and a[1]

(3.1.3.11)

(3.1.3.12)

Using Maples dsolve command to solve the given deq.

(3.1.3.13)

NOTE: Since this deq has NO singular points the radius of convergence for this solution is ∞. ie it is valid for all values of x.

Example 2. Solve the DEQ y'' + xy' +(2x-3)y = 0 near x = -1.

Note x = -1 is an ordinary point for this deq.

We first make the substitution t = x - (-1) = x + 1. The resulting deq can then be solved near t = 0 using the above techniques.

ie Since y(x) = y(t-1) and $Typesetting:-mrow(Typesetting:-mfrac(Typesetting:-mi($ and $Typesetting:-mrow(Typesetting:-mfrac(Typesetting:-mrow(Typesetting:-msup(Typesetting:-mi($ the eqn becomes $Typesetting:-mrow(Typesetting:-mfrac(Typesetting:-mrow(Typesetting:-msup(Typesetting:-mi($

First create the power series solution

To calculate the derivative of Y(x)

(3.1.4.1)

(3.1.4.2)

converting the coefficients into power series

Combining these series as defined by the equation

Using the op command to identify the coefficients , setting them equal to zero, and solving for them in terms of a[0] and a[1]

Substituting the coeff into Y(x) and converting the solution into a polynomial

Substituting that t = x+1 we obtain the solution to the original problem

Again since this eqn has NO singular points this solution is valid for all values of x.

Given initial conditions we could then proceed to determine the values for and

Suppose that y(-1) = 2 and y'(-1) = -2

(3.1.4.7)

(3.1.4.8)

Using Maple's dsolve command.

Solutions about a regular singular point --- Method of Frobenius.

Definition 3. If is a regular singular point of y'' + p(x) y' +q(x) y = 0, then the indicial equation for this point is

whereand

The roots of the indicial equation are called the exponents(indices) of the singularity

Frobenius method of solving ordinary differential equations near a regular singular point, , by positing a solution of the form

the values of r and the coefficients are then found by iteration by substituting the potential solution into the equation.

NOTE The first step in this method is to find the roots and (Re ) of the indicial equation. Then utilizing the larger root Frobenius's theorem assures us that our deq has a series solution of the form above and that this series converges for all x such thatwhere R is the distance from to the nearest other singular point (real or complex).

If is not an integer, then there exist two linearly independent solutions of the form

Ifthen there exist two linearly independent solutions of the form

If is a positive integer, then there exist two linearly independent solutions of the form

where C is a constant that could be zero.

Example 1. Find a series solution for the differential equation

Now x = 0 and x = -2 are both singular points for this deq.

Also x = 0 is a regular singular point since and are analytic at

We are looking for a solution of the form

Step 1. Define our deq

(3.2.1.1)

Step 2. Define our series solution

Step 3 . Substitute our series into the deq.

Step 4. Determine the coefficient of the term ( it in effect is the indicial eqn needed to determine the values of r in Frobenius' method )

(3.2.1.6)

From the above we see that r = 1 or r = 1/2

Step 5. We use the larger value of r = 1 first. We now substitute that y1(0) = a , D(y1)(0) = b and r =1 into the coefficients of the lhs our eqn1

We now want to solve the above coefficients = 0 for the values of the derivatives in each of themTo do this we must drop the first and 2nd items. NOTE : the 2nd item a + 3b =0 says that b = -1/3 a

Step 6 . Substitute these values into slna minus the last three terms since we have not calculated their correct coefficients

(3.2.1.10)

It may be possible to obtain the second independent series solution by repeating the above using the other value of r ( This will work in this example since the two r values do not differ by an integer).

We now want to solve the above coefficients = 0 for the values of the derivatives in each of themTo do this we must drop the first and 2nd items. NOTE : 2nd item says b=-3/4a

Finally we substitute these values into slna minus the last three terms since we have not calculated their correct coefficients

(3.2.1.14)

Hence the general solution to the given differential equation will be

MAPLE dsolve solution

(3.2.1.16)

Example 2. Find a series solution for the differential equation about the point x = 0

x = 0 is a regular singular point for this deq.

(3.2.2.1)

(3.2.2.5)

We see that r = 1 is a double root and proceed to find the first solution using the procedures outlined above.

(3.2.2.9)

We now proceed to find the second solution using the fact that it will be of the form

(3.2.2.10)

(3.2.2.11)

Since y1(x) is a solution of the deq the factor on ln(x) is zero ( We also convert our series into a finite sum in order to deal with the summations...)

(3.2.2.14)

(3.2.2.15)

(3.2.2.19)

MAPLE dsolve solution.....

(3.2.2.20)

Example 3. Find a series solution for the differential equation

Note x = 0 is a regular singular point for this deq.

(3.2.3.1)

(3.2.3.4)

roots are 0 and -3 In this case they differ by a positive integer

roots are r = 0 and -3. Using the larger root r = 0

(3.2.3.8)

A second solution of the form and where C is a constant that may be zero.

(3.2.3.9)

Again since y1(x) is a solution the factor on ln(x) is 0.

(3.2.3.12)

(3.2.3.13)

(3.2.3.14)

NOTE The factor on for another independent solution we must choose to be nonzero.

In effect sln2aff is a general solution to the given deq.

(3.2.3.15)

Legal Notice: The copyright for this application is owned by the author(s). Neither Maplesoft nor the author are responsible for any errors contained within and are not liable for any damages resulting from the use of this material. This application is intended for non-commercial, non-profit use only. Contact the author for permission if you wish to use this application in for-profit activities.