Differential Equations Project

Computer Lab Modules

 

Power Series Solution of O.D.E.'s

 

 

Background. Power series can be used to solve differential equations at an ordinary point.

 

Computer Lab Work.

Exercise 1. Use series to solve the D. E. [Graphics:e21.txtgr1.gif].

Solution. Form the substitution series and the equation to solve.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr2.gif]

Form the set of equations to solve and do it.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr4.gif]

Two linearly independent series can be obtained.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr5.gif]

After you are done, use Mathematica's DSolve subroutine to get the answer and check out its series expansion.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr6.gif]

At this time we could plot the series approximations and the analytic solutions.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr7.gif]
[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr8.gif]

If we look at the series in more depth we will be able to obtain the analytic solutions as infinite sums.
First find the recursive formula for the coefficients of [Graphics:e21.txtgr9.gif].

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr10.gif]

Now look at each series individually. This requires a lot a hand work.

For the first series we have c[0] = 1 , c[1] = 0.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr11.gif]

It should be evident that all the odd terms compute to be zero.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr12.gif]

The even terms are:

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr13.gif]

Use the recursive formula [Graphics:e21.txtgr14.gif] and some quotes " " to get things to print out nice so that we can recognize the pattern.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr15.gif]

For the series of even terms the coefficients of [Graphics:e21.txtgr16.gif] can be seen to be [Graphics:e21.txtgr17.gif] where the recursive formula is:

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr18.gif]

Now we can add up all the terms for the function f1[x].

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr19.gif]


Now look at the second series where we have c[0] = 0 , c[1] = 1.

It should be evident that all the even terms compute to be zero.

The odd terms do not compute to be zero.

Use the recursive formula [Graphics:e21.txtgr20.gif] and some quotes " " to get things to print out nice so that we can recognize the pattern.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr21.gif]

For the series of odd terms the coefficients of [Graphics:e21.txtgr22.gif] can be seen to be [Graphics:e21.txtgr23.gif] where the recursive formula is:

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr24.gif]

Now we can add up all the terms for the function f2[x].

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr25.gif]


Exercise 2. Use series to solve the D. E. [Graphics:e21.txtgr26.gif]

Solution. Form the substitution series and the equation to solve.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr27.gif]

Form the set of equations to solve and do it.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr28.gif]

Two linearly independent series can be obtained.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr29.gif]

After you are done, use Mathematica's DSolve subroutine to get the answer and check out its series expansion.
You will need to supply initial conditions and solve two I.V.P.'s

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr30.gif]

At this time we could plot the series approximations and the analytic solutions.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr31.gif]
[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr32.gif]

If we look at the series in more depth we will be able to obtain the analytic solutions as infinite sums.
First find the recursive formula for the coefficients of [Graphics:e21.txtgr33.gif].

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr34.gif]

Now look at each series individually. This requires a lot a hand work.

Go back to the solutions and see that every third term is zero starting with c[2] = 0.

For the first series we have c[0] = 1 , c[1] = 0.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr35.gif]

It should be evident that all the terms c[1], c[4], c[7],... compute to be zero.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr36.gif]

It should be evident that all the terms c[2], c[5], c[8],... compute to be zero.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr37.gif]

The terms starting with c[0] are:

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr38.gif]

The following quotients are used in the construction of the terms. Do you see the simplification ?

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr39.gif]

For the series of terms the coefficients of [Graphics:e21.txtgr40.gif] are [Graphics:e21.txtgr41.gif] where the recursive formula is:

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr42.gif]

Now we can add up all the terms for the function f1[x].

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr43.gif]


Now look at the second series where we have c[0] = 0 , c[1] = 1.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr44.gif]

It should be evident that all the terms c[0], c[3], c[0],... compute to be zero.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr45.gif]

It should be evident that all the terms c[2], c[5], c[8],... compute to be zero.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr46.gif]

The terms starting with c[1] are:

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr47.gif]

For the series of terms the coefficients of [Graphics:e21.txtgr48.gif] are [Graphics:e21.txtgr49.gif] where the recursive formula is:

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr50.gif]

Now we can add up all the terms for the function f1[x].

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr51.gif]


Exercise 3. Use series to solve Bessel's D. E. [Graphics:e21.txtgr52.gif]

Solution. Form the substitution series and the equation to solve.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr53.gif]

Form the set of equations to solve and do it.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr54.gif]

For this differential equation it is known that there is only one power series solution.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr55.gif]

The second solution is not a simple power series.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr56.gif]
[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr57.gif]

The function BesselK[0, I x] is NOT a power series.

After you are done, use Mathematica's DSolve subroutine to get the answer and check out its series expansion.
You will need to supply initial conditions and solve two I.V.P.'s

At this time we could plot the series approximations and the analytic solutions.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr58.gif]

If we look at the series in more depth we will be able to obtain the analytic solutions as infinite sums.
First find the recursive formula for the coefficients of [Graphics:e21.txtgr59.gif].

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr60.gif]

Now look at each series individually. This requires a lot a hand work.

For the first series we have c[0] = 1 , c[1] = 0.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr61.gif]
[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr62.gif]

For the series of terms the coefficients of [Graphics:e21.txtgr63.gif] are [Graphics:e21.txtgr64.gif] where the recursive formula is:

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr65.gif]

Now we can add up all the terms for the function f1[x].

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr66.gif]


Exercise 4. Use series to solve the D. E. [Graphics:e21.txtgr67.gif].

Solution. Form the substitution series and the equation to solve.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr68.gif]

Form the set of equations to solve and do it.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr69.gif]

After you are done, use Mathematica's DSolve subroutine to get the answer and check out its series expansion.

[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr70.gif]
[Graphics:e21.txtgr3.gif][Graphics:e21.txtgr71.gif]
 

Solutions.

 

 

Return to the Differential Equations Project

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(c) John H. Mathews, 1998