# 17.4: Series Solutions of Differential Equations

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Learning Objectives

• Use power series to solve first-order and second-order differential equations.

Previously, we studied how functions can be represented as power series, (y(x)=sum_{n=0}^{infty} a_nx^n). We also saw that we can find series representations of the derivatives of such functions by differentiating the power series term by term. This gives

[y′(x)=sum_{n=1}^{infty}na_nx^{n−1} onumber]

and

[y″(x)=sum_{n=2}^{infty}n(n−1)a_nx^{n−2}. onumber]

In some cases, these power series representations can be used to find solutions to differential equations.

The examples and exercises in this section were chosen for which power solutions exist. However, it is not always the case that power solutions exist. Those of you interested in a more rigorous treatment of this topic should review the differential equations section of the.

Problem-Solving Strategy: Finding Power Series Solutions to Differential Equations

1. Assume the differential equation has a solution of the form [y(x)=sum_{n=0}^{infty}a_nx^n.]
2. Differentiate the power series term by term to get [y′(x)=sum_{n=1}^{infty}na_nx^{n−1}]and[y″(x)=sum_{n=2}^{infty}n(n−1)a_nx^{n−2}.]
3. Substitute the power series expressions into the differential equation.
4. Re-index sums as necessary to combine terms and simplify the expression.
5. Equate coefficients of like powers of (x) to determine values for the coefficients (a_n) in the power series.
6. Substitute the coefficients back into the power series and write the solution.

Example (PageIndex{1}): Series Solutions to Differential Equations

Find a power series solution for the following differential equations.

1. (y''−y=0)
2. ((x^2−1)y″+6xy′+4y=−4)

Solution

Part a

Assume

[y(x)=sum_{n=0}^{∞}a_nx^n ag{step 1}]

Then,

[y′(x)=sum_{n=1}^{∞}na_nx^{n−1} ag{step 2A}]

and

[y″(x)=sum_{n=2}^{∞}n(n−1)a_nx^{n−2} ag{step 2B}]

We want to find values for the coefficients (a_n) such that

[egin{align} y″−y =0 onumber [4pt] sum_{n=2}^{∞}n(n−1)a_nx^{n−2}−sum_{n=0}^{∞}a_nx^n onumber =0 ag{step 3}. end{align}]

We want the indices on our sums to match so that we can express them using a single summation. That is, we want to rewrite the first summation so that it starts with (n=0).

To re-index the first term, replace (n) with (n+2) inside the sum, and change the lower summation limit to (n=0.) We get

[sum_{n=2}^{∞}n(n−1)a_nx^{n−2}=sum_{n=0}^∞ (n+2)(n+1)a_{n+2}x^n. onumber]

This gives

[egin{align}sum_{n=0}^{∞}(n+2)(n+1)a_{n+2}x^n−sum_{n=0}^{∞}a_nx_n =0 onumber [4pt] sum_{n=0}^{∞}[(n+2)(n+1)a_{n+2}−a_n]x^n =0 ag{step 4}.end{align}]

Because power series expansions of functions are unique, this equation can be true only if the coefficients of each power of (x) are zero. So we have

[(n+2)(n+1)a_{n+2}−a_n=0 ext{ for }n=0,1,2,…. onumber]

This recurrence relationship allows us to express each coefficient (a_n) in terms of the coefficient two terms earlier. This yields one expression for even values of (n) and another expression for odd values of (n). Looking first at the equations involving even values of (n), we see that

[egin{align*}a_2 = dfrac{a_0}{2} a_4 = dfrac{a_2}{4⋅3} = dfrac{a_0}{4!} a_6 = dfrac{a_4}{6⋅5} =dfrac{a_0}{6!} ; ⋮. end{align*}]

Thus, in general, when n is even,

[a_n=dfrac{a_0}{n!}. ag{step 5}]

For the equations involving odd values of n, we see that

[egin{align*}a_3 =dfrac{a_1}{3⋅2}=dfrac{a_1}{3!} a_5 = dfrac{a_3}{5⋅4}=dfrac{a_1}{5!} a_7 = dfrac{a_5}{7⋅6}=dfrac{a_1}{7!} ; ⋮. end{align*}]

Therefore, in general, when n is odd,

[a_n=dfrac{a_1}{n!}. ag{step 5}]

Putting this together, we have

[egin{align*}y(x) &= sum_{n=0}^{∞}a_nx^n [4pt] &=a_0+a_1x+dfrac{a_0}{2}x^2+dfrac{a_1}{3!}x^3+dfrac{a_0}{4!}x^4+dfrac{a_1}{5!}x^5+⋯. end{align*}]

Re-indexing the sums to account for the even and odd values of n separately, we obtain

[y(x)=a_0 sum{k=0}^{∞} dfrac{1}{(2k)!}x^{2k}+a_1 sum_{k=0}^{∞}dfrac{1}{(2k+1)!}x^{2k+1}. ag{step 6}]

Analysis for part a.

As expected for a second-order differential equation, this solution depends on two arbitrary constants. However, note that our differential equation is a constant-coefficient differential equation, yet the power series solution does not appear to have the familiar form (containing exponential functions) that we are used to seeing. Furthermore, since (y(x)=c_1e^x+c_2e^{−x}) is the general solution to this equation, we must be able to write any solution in this form, and it is not clear whether the power series solution we just found can, in fact, be written in that form.

Fortunately, after writing the power series representations of (e^x) and (e^{−x},) and doing some algebra, we find that if we choose

[c_0=dfrac{(a_0+a_1)}{2}, c_1=dfrac{(a_0−a_1)}{2}, onumber]

we then have (a_0=c_0+c_1) and (a_1=c_0−c_1,) and

[egin{align*}y(x) &= a_0+a_1x+dfrac{a_0}{2}x^2+dfrac{a_1}{3!}x^3+dfrac{a_0}{4!}x^4+dfrac{a_1}{5!}x^5+⋯ [4pt] &=(c_0+c_1)+(c_0−c_1)x+dfrac{(c_0+c_1)}{2}x^2+dfrac{(c_0−c_1)}{3!}x^3+dfrac{(c_0+c_1)}{4!}x^4+dfrac{(c_0−c_1)}{5!}x^5+⋯[4pt] &=c_0 sum_{n=0}^{∞} dfrac{x^n}{n!}+c_1 sum_{n=0}^{∞}dfrac{(−x)^n}{n!} [4pt] &=c_0e^x+c_1e^{−x}.end{align*}]

So we have, in fact, found the same general solution. Note that this choice of (c_1) and (c_2) is not obvious. This is a case when we know what the answer should be, and have essentially “reverse-engineered” our choice of coefficients.

Part b

Assume

[y(x)=sum_{n=0}^{∞}a_nx^n ag{step 1}]

Then,

[y′(x)=sum_{n=1}^{∞}na_nx^{n−1} ag{step 2}]

and

[y″(x)=sum_{n=2}^{∞}n(n−1)a_nx^{n−2} ag{step 2}]

We want to find values for the coefficients (a_n) such that

[egin{align*}(x^2−1)y″+6xy′+4y &=−4 (x^2−1) sum_{n=2}^{∞}n(n−1)a_nx^{n−2}+6x sum_{n=1}^{∞}na_nx^{n−1}+4 sum_{n=0}^{∞}a_nx^n &=−4 [4pt] x^2 sum_{n=2}^{∞} n(n−1)a_nx^{n−2}−sum_{n=2}^{∞}n(n−1)a_nx^{n−2}+6x sum_{n=1}^{∞}na_nx^{n−1}+4 sum_{n=0}^{∞}a_nx^n &=−4. end{align*}]

Taking the external factors inside the summations, we get

[sum_{n=2}^{∞}n(n−1)a_nx^n−sum_{n=2}^{∞}n(n−1)a_nx^{n−2}+sum_{n=1}^∞ 6na_nx^n+ sum_{n=0}^∞ 4a_nx^n=−4 ag{step 3}. ]

Now, in the first summation, we see that when (n=0) or (n=1), the term evaluates to zero, so we can add these terms back into our sum to get

[sum_{n=2}^{∞}n(n−1)a_nx^n=sum_{n=0}^∞ n(n−1)a_nx^n. onumber]

Similarly, in the third term, we see that when (n=0), the expression evaluates to zero, so we can add that term back in as well. We have

[sum_{n=1}^∞ 6na_nx^n=sum_{n=0}^∞6na_nx^n. onumber]

Then, we need only shift the indices in our second term. We get

[sum_{n=2}^∞n(n−1)a_nx^{n−2}=sum_{n=0}^∞(n+2)(n+1)a_{n+2}x^n. onumber]

Thus, we have

[egin{align*} sum_{n=0}^∞n(n−1)a_nx^n−sum_{n=0}^∞(n+2)(n+1)a_{n+2}x^n+sum_{n=0}^∞6na_nx^n+sum_{n=0}^∞4a_nx^n &=−4 ag{step 4} [4pt] sum_{n=0}^∞[n(n−1)a_n−(n+2)(n+1)a_{n+2}+6na_n+4a_n]x^n &=−4 [4pt] sum_{n=0}^∞[(n^2−n)a_n+6na_n+4a_n−(n+2)(n+1)a_{n+2}]x^n &=−4 [4pt] sum_{n=0}^∞[n^2a_n+5na_n+4a_n−(n+2)(n+1)a_{n+2}]x^n &=−4 [4pt] sum_{n=0}^∞ [(n^2+5n+4)a_n−(n+2)(n+1)a_{n+2}]x^n &=−4 [4pt] sum_{n=0}^∞[(n+4)(n+1)a_n−(n+2)(n+1)a_{n+2}]x^n &=−4 end{align*}]

Looking at the coefficients of each power of (x), we see that the constant term must be equal to (−4), and the coefficients of all other powers of x must be zero. Then, looking first at the constant term,

[egin{aligned}4a_0−2a_2 &=−4 a_2 &=2a_0+2 end{aligned} ag{step 3} ]

For (n≥1), we have

[egin{align*}(n+4)(n+1)a_n−(n+2)(n+1)a_{n+2} &= 0 [4pt] (n+1)[(n+4)a_n−(n+2)a_{n+2}] &=0. end{align*}]

Since (n≥1, n+1≠0,) we see that

[(n+4)a_n−(n+2)a_{n+2}=0 onumber]

and thus

[a_{n+2}=dfrac{n+4}{n+2}a_n. onumber]

For even values of (n), we have

[egin{align*} a_4 =dfrac{6}{4}(2a_0+2)=3a_0+3 a_6 = dfrac{8}{6}(3a_0+3)=4a_0+4 ; ⋮. end{align*}]

In general,

[a_{2k}=(k+1)(a_0+1). ag{step 5} ]

For odd values of n, we have

[egin{align*}a_3 =dfrac{5}{3}a_1 a_5 = dfrac{7}{5}a_3=dfrac{7}{3}a_1 a_7 =dfrac{9}{7}a_5=dfrac{9}{3} a_1=3a_1 ; ⋮. end{align*}]

In general,

[a_{2k+1}=dfrac{2k+3}{3}a_1. ag{step 5 continued}]

Putting this together, we have

[y(x)=sum{k=0}^∞ (k+1)(a_0+1)x^{2k}+sum_{k=0}^∞ (dfrac{2k+3}{3})a_1x^{2k+1}. ag{step 6}]

Exercise (PageIndex{1})

Find a power series solution for the following differential equations.

1. (y′+2xy=0)
2. ((x+1)y′=3y)
Hint

(y(x)=a_0 sum_{n=0}^∞ dfrac{(−1)^n}{n!}x^{2n}=a_0e^{−x^2})

(y(x)=a_0(x+1)^3)

## Bessel functions

We close this section with a brief introduction to Bessel functions. Complete treatment of Bessel functions is well beyond the scope of this course, but we get a little taste of the topic here so we can see how series solutions to differential equations are used in real-world applications. The Bessel equation of order n is given by

[x^2y″+xy′+(x^2−n^2)y=0.]

This equation arises in many physical applications, particularly those involving cylindrical coordinates, such as the vibration of a circular drum head and transient heating or cooling of a cylinder. In the next example, we find a power series solution to the Bessel equation of order 0.

Example (PageIndex{2}): Power Series Solution to the Bessel Equation

Find a power series solution to the Bessel equation of order 0 and graph the solution.

Solution

The Bessel equation of order 0 is given by

[x^2y″+xy′+x^2y=0. onumber]

We assume a solution of the form (y=sum_{n=0}^∞ a_nx^n). Then (y′(x)=sum_{n=1}^∞ na_nx^{n−1}) and (y''(x)=sum_{n=2}^∞n(n−1)a_nx^{n−2}.)Substituting this into the differential equation, we get

[egin{array}{ll} x^2 sum_{n=2}^∞ n(n−1)a_nx^{n−2}+x sum_{n=1}^∞ na_nx^{n−1}+x^2 sum_{n=0}^∞ a_nx^n=0 ext{Substitution.} sum_{n=2}^∞ n(n−1)a_nx^n+sum_{n=1}^∞ na_nx^n+ sum_{n=0}^∞ a_nx^{n+2}=0 ext{Bring external factors within sums.} sum_{n=2}^∞ n(n−1)a_nx^n+sum_{n=1}^∞ na_nx^n+sum_{n=2}^∞ a_{n−2}x^n=0 ext{Re-index third sum.} sum_{n=2}^∞ n(n−1)a_nx^n+a_1x+sum_{n=2}^∞ na_nx^n+sum_{n=2}^∞ a_{n−2}x^n=0 ext{Separate }n=1 ext{ term from second sum.} a_1x+sum_{n=2}^∞ [n(n−1)a_n+na_n+a_{n−2}]x^n=0 ext{Collect summation terms.} a_1x+sum_{n=2}^∞ [(n^2−n)a_n+na_n+a_{n−2}]x^n=0 ext{Multiply through in first term.} a_1x+sum_{n=2}^∞ [n^2a_n+a_{n−2}]x^n=0. ext{Simplify.} end{array}]

Then, (a_1=0), and for (n≥2,)

[egin{align*} n^2a_n+a_{n−2} = 0 a_n=−dfrac{1}{n^2}a_{n−2}. end{align*} ]

Because (a_1=0), all odd terms are zero. Then, for even values of n, we have

[egin{align*}a_2 =−dfrac{1}{2^2}a_0 a_4 = −dfrac{1}{4^2}a_2=dfrac{1}{4^2⋅2^2} a_0. a_6 =−dfrac{1}{6^2}a_4 =−dfrac{1}{6^2⋅4^2⋅2^2}a_0 end{align*}]

In general,

[a_{2k}=dfrac{(−1)^k}{(2)^{2k}(k!)^2}a_0. onumber]

Thus, we have

[y(x)=a_0 sum_{k=0}^∞ dfrac{(−1)^k}{(2)^{2k}(k!)^2}x^{2k}. onumber]

The graph appears below.

Exercise (PageIndex{2})

Verify that the expression found in Example (PageIndex{2}) is a solution to the Bessel equation of order 0.

Hint

Differentiate the power series term by term and substitute it into the differential equation.

## Key Concepts

• Power series representations of functions can sometimes be used to find solutions to differential equations.
• Differentiate the power series term by term and substitute into the differential equation to find relationships between the power series coefficients.

## Series Solutions in MATLAB 2020a and later

As of MATLAB 2020a, the ability to request series solutions to differential equations using dsolve now exists, but the syntax is slightly different from what we guessed it would be when the 2019 edition of Differential Equations with MATLAB was written. On this page, we explain the correct syntax and give some actual examples.

To request a series solution to a differential equation using dsolve, begin with the ordinary dsolve code, but add 'ExpansionPoint' followed by the point around which one wants a series solution. Usually this will be the point at which the initial condition is specified. Specify 'Order' to change the number of terms in the series, just as you would with the series command. We give a number of examples.

&diams Solving y' = y with initial condition y(0)=1. The solution is the exponential function. syms y(t) dsolve(diff(y)==y, y(0)==1, 'ExpansionPoint', 0) This produces the output ans = t^5/120 + t^4/24 + t^3/6 + t^2/2 + t + 1

Suppose a2 is nonzero for all z. Then we can divide throughout to obtain

The power series method calls for the construction of a power series solution

If a2 is zero for some z, then the Frobenius method, a variation on this method, is suited to deal with so called "singular points". The method works analogously for higher order equations as well as for systems.

We can try to construct a series solution

Substituting these in the differential equation

Making a shift on the first sum

If this series is a solution, then all these coefficients must be zero, so for both k=0 and k>0:

We can rearrange this to get a recurrence relation for Ak+2.

We can determine A0 and A1 if there are initial conditions, i.e. if we have an initial value problem.

and the series solution is

which we can break up into the sum of two linearly independent series solutions:

which can be further simplified by the use of hypergeometric series.

A much simpler way of solving this equation (and power series solution in general) using the Taylor series form of the expansion. Here we assume the answer is of the form

If we do this, the general rule for obtaining the recurrence relationship for the coefficients is

In this case we can solve the Hermite equation in fewer steps:

The power series method can be applied to certain nonlinear differential equations, though with less flexibility. A very large class of nonlinear equations can be solved analytically by using the Parker–Sochacki method. Since the Parker–Sochacki method involves an expansion of the original system of ordinary differential equations through auxiliary equations, it is not simply referred to as the power series method. The Parker–Sochacki method is done before the power series method to make the power series method possible on many nonlinear problems. An ODE problem can be expanded with the auxiliary variables which make the power series method trivial for an equivalent, larger system. Expanding the ODE problem with auxiliary variables produces the same coefficients (since the power series for a function is unique) at the cost of also calculating the coefficients of auxiliary equations. Many times, without using auxiliary variables, there is no known way to get the power series for the solution to a system, hence the power series method alone is difficult to apply to most nonlinear equations.

The power series method will give solutions only to initial value problems (opposed to boundary value problems), this is not an issue when dealing with linear equations since the solution may turn up multiple linearly independent solutions which may be combined (by superposition) to solve boundary value problems as well. A further restriction is that the series coefficients will be specified by a nonlinear recurrence (the nonlinearities are inherited from the differential equation).

In order for the solution method to work, as in linear equations, it is necessary to express every term in the nonlinear equation as a power series so that all of the terms may be combined into one power series.

As an example, consider the initial value problem

which describes a solution to capillary-driven flow in a groove. There are two nonlinearities: the first and second terms involve products. The initial values are given at η = 1 , which hints that the power series must be set up as:

which makes the initial values very easy to evaluate. It is necessary to rewrite the equation slightly in light of the definition of the power series,

so that the third term contains the same form η − 1 that shows in the power series.

The last consideration is what to do with the products substituting the power series in would result in products of power series when it's necessary that each term be its own power series. This is where the Cauchy product

is useful substituting the power series into the differential equation and applying this identity leads to an equation where every term is a power series. After much rearrangement, the recurrence

∑ j = 0 i ( ( j + 1 ) ( j + 2 ) c i − j c j + 2 + 2 ( i − j + 1 ) ( j + 1 ) c i − j + 1 c j + 1 ) + i c i + ( i + 1 ) c i + 1 = 0 ^left((j+1)(j+2)c_c_+2(i-j+1)(j+1)c_c_ ight)+ic_+(i+1)c_=0>

## 50 Series Solutions of Differential Equations

In Introduction to Power Series, we studied how functions can be represented as power series, We also saw that we can find series representations of the derivatives of such functions by differentiating the power series term by term. This gives and In some cases, these power series representations can be used to find solutions to differential equations.

Be aware that this subject is given only a very brief treatment in this text. Most introductory differential equations textbooks include an entire chapter on power series solutions. This text has only a single section on the topic, so several important issues are not addressed here, particularly issues related to existence of solutions. The examples and exercises in this section were chosen for which power solutions exist. However, it is not always the case that power solutions exist. Those of you interested in a more rigorous treatment of this topic should consult a differential equations text.

1. Assume the differential equation has a solution of the form
2. Differentiate the power series term by term to get and
3. Substitute the power series expressions into the differential equation.
4. Re-index sums as necessary to combine terms and simplify the expression.
5. Equate coefficients of like powers of to determine values for the coefficients in the power series.
6. Substitute the coefficients back into the power series and write the solution.

Find a power series solution for the following differential equations.

1. Assume (step 1). Then, and (step 2). We want to find values for the coefficients such that

We want the indices on our sums to match so that we can express them using a single summation. That is, we want to rewrite the first summation so that it starts with
To re-index the first term, replace n with inside the sum, and change the lower summation limit to We get

Because power series expansions of functions are unique, this equation can be true only if the coefficients of each power of x are zero. So we have

This recurrence relationship allows us to express each coefficient in terms of the coefficient two terms earlier. This yields one expression for even values of n and another expression for odd values of n. Looking first at the equations involving even values of n, we see that

Thus, in general, when n is even, (step 5).
For the equations involving odd values of n, we see that

Therefore, in general, when n is odd, (step 5 continued).
Putting this together, we have

Re-indexing the sums to account for the even and odd values of n separately, we obtain

Analysis for part a.
As expected for a second-order differential equation, this solution depends on two arbitrary constants. However, note that our differential equation is a constant-coefficient differential equation, yet the power series solution does not appear to have the familiar form (containing exponential functions) that we are used to seeing. Furthermore, since is the general solution to this equation, we must be able to write any solution in this form, and it is not clear whether the power series solution we just found can, in fact, be written in that form.
Fortunately, after writing the power series representations of and and doing some algebra, we find that if we choose

we then have and and

Taking the external factors inside the summations, we get

Now, in the first summation, we see that when or the term evaluates to zero, so we can add these terms back into our sum to get

Similarly, in the third term, we see that when the expression evaluates to zero, so we can add that term back in as well. We have

Then, we need only shift the indices in our second term. We get

Looking at the coefficients of each power of x, we see that the constant term must be equal to and the coefficients of all other powers of x must be zero. Then, looking first at the constant term,

For we have

Since we see that

For even values of n, we have

In general, (step 5).
For odd values of n, we have

In general, (step 5 continued).
Putting this together, we have

Find a power series solution for the following differential equations.

We close this section with a brief introduction to Bessel functions . Complete treatment of Bessel functions is well beyond the scope of this course, but we get a little taste of the topic here so we can see how series solutions to differential equations are used in real-world applications. The Bessel equation of order n is given by

This equation arises in many physical applications, particularly those involving cylindrical coordinates, such as the vibration of a circular drum head and transient heating or cooling of a cylinder. In the next example, we find a power series solution to the Bessel equation of order 0.

Find a power series solution to the Bessel equation of order 0 and graph the solution.

The Bessel equation of order 0 is given by

We assume a solution of the form Then and Substituting this into the differential equation, we get

Then, and for

Because all odd terms are zero. Then, for even values of n, we have

Verify that the expression found in (Figure) is a solution to the Bessel equation of order 0.

Differentiate the power series term by term and substitute it into the differential equation.

### Key Concepts

• Power series representations of functions can sometimes be used to find solutions to differential equations.
• Differentiate the power series term by term and substitute into the differential equation to find relationships between the power series coefficients.

Find a power series solution for the following differential equations.

The differential equation is a Bessel equation of order 1. Use a power series of the form to find the solution.

### Chapter Review Exercises

True or False? Justify your answer with a proof or a counterexample.

If and are both solutions to then is also a solution.

The following system of algebraic equations has a unique solution:

is a solution to the second-order differential equation

To find the particular solution to a second-order differential equation, you need one initial condition.

Classify the differential equation. Determine the order, whether it is linear and, if linear, whether the differential equation is homogeneous or nonhomogeneous. If the equation is second-order homogeneous and linear, find the characteristic equation.

second order, linear, homogeneous,

first order, nonlinear, nonhomogeneous

For the following problems, find the general solution.

For the following problems, find the solution to the initial-value problem, if possible.

For the following problems, find the solution to the boundary-value problem.

For the following problem, set up and solve the differential equation.

The motion of a swinging pendulum for small angles can be approximated by where is the angle the pendulum makes with respect to a vertical line, g is the acceleration resulting from gravity, and L is the length of the pendulum. Find the equation describing the angle of the pendulum at time assuming an initial displacement of and an initial velocity of zero.

The following problems consider the “beats” that occur when the forcing term of a differential equation causes “slow” and “fast” amplitudes. Consider the general differential equation that governs undamped motion. Assume that

Find the general solution to this equation (Hint: call ).

Assuming the system starts from rest, show that the particular solution can be written as

[T] Using your solutions derived earlier, plot the solution to the system over the interval Find, analytically, the period of the fast and slow amplitudes.

For the following problem, set up and solve the differential equations.

An opera singer is attempting to shatter a glass by singing a particular note. The vibrations of the glass can be modeled by where represents the natural frequency of the glass and the singer is forcing the vibrations at For what value would the singer be able to break that glass? (Note: in order for the glass to break, the oscillations would need to get higher and higher.)

## 17.4: Series Solutions of Differential Equations

The purpose of this section is not to do anything new with a series solution problem. Instead it is here to illustrate that moving into a higher order differential equation does not really change the process outside of making it a little longer.

Back in the Series Solution chapter we only looked at 2 nd order differential equations so we’re going to do a quick example here involving a 3 rd order differential equation so we can make sure and say that we’ve done at least one example with an order larger than 2.

Recall that we can only find a series solution about ( = 0) if this point is an ordinary point, or in other words, if the coefficient of the highest derivative term is not zero at ( = 0). That is clearly the case here so let’s start with the form of the solutions as well as the derivatives that we’ll need for this solution.

Plugging into the differential equation gives,

Now, move all the coefficients into the series and do appropriate shifts so that all the series are in terms of ().

Next, let’s notice that we can start the second series at (n = 1) since that term will be zero. So let’s do that and then we can combine the second and third terms to get,

So, we got a nice simplification in the new series that will help with some further simplification. The new second series can now be started at (n = 0) and then combined with the first series to get,

We can now set the coefficients equal to get a fairly simply recurrence relation.

Solving the recurrence relation gives,

Now we need to start plugging in values of (n) and this will be one of the main areas where we can see a somewhat significant increase in the amount of work required when moving into a higher order differential equation.

( displaystyle n = 0:hspace<0.25in> = 0)

Okay, we can now break the coefficients down into 4 sub cases given by (>), (>), (>) and (>) for (k = 0,1,2,3, ldots ) We’ll give a semi-detailed derivation for (>) and then leave the rest to you with only couple of comments as they are nearly identical derivations.

First notice that all the (>) terms have () in them and they will alternate in sign. Next notice that we can turn the denominator into a factorial, (left( <4k> ight)!) to be exact, if we multiply top and bottom by the numbers that are already in the numerator and so this will turn these numbers into squares. Next notice that the product in the top will start at 1 and increase by 4 until we reach (4k - 3). So, taking all of this into account we get,

and notice that this will only work starting with (k = 1) as we won’t get () out of this formula as we should by plugging in (k = 0).

Now, for (>) the derivation is almost identical and so the formula is,

and again notice that this won’t work for (k = 0)

The formula for (>) is again nearly identical except for this one note that we also need to multiply top and bottom by a 2 in order to get the factorial to appear in the denominator and so the formula here is,

noticing yet one more time that this won’t work for (k = 0).

Finally, we have (> = 0) for (k = 0,1,2,3, ldots )

Now that we have all the coefficients let’s get the solution,

Collecting up the terms that contain the same coefficient (except for the first one in each case since the formula won’t work for those) and writing everything as a set of series gives us our solution,

So, there we have it. As we can see the work in getting formulas for the coefficients was a little messy because we had three formulas to get, but individually they were not as bad as even some of them could be when dealing with 2 nd order differential equations. Also note that while we got lucky with this problem and we were able to get general formulas for the terms the higher the order the less likely this will become.

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## 17.4: Series Solutions of Differential Equations

We have seen how to solve a restricted collection of differential equations, or more accurately, how to attempt to solve them&mdashwe may not be able to find the required anti-derivatives. Not surprisingly, non-linear equations can be even more difficult to solve. Yet much is known about solutions to some more general equations.

Suppose $phi(t,y)$ is a function of two variables. A more general class of first order differential equations has the form $dsdot y=phi(t,y)$. This is not necessarily a linear first order equation, since $phi$ may depend on $y$ in some complicated way note however that $dsdot y$ appears in a very simple form. Under suitable conditions on the function $phi$, it can be shown that every such differential equation has a solution, and moreover that for each initial condition the associated initial value problem has exactly one solution. In practical applications this is obviously a very desirable property.

Example 17.4.1 The equation $dsdot y=t-y^2$ is a first order non-linear equation, because $y$ appears to the second power. We will not be able to solve this equation.

Example 17.4.2 The equation $dsdot y=y^2$ is also non-linear, but it is separable and can be solved by separation of variables.

Not all differential equations that are important in practice can be solved exactly, so techniques have been developed to approximate solutions. We describe one such technique, Euler's Method, which is simple though not particularly useful compared to some more sophisticated techniques.

Suppose we wish to approximate a solution to the initial value problem $dsdot y=phi(t,y)$, $ds y(t_0)=y_0$, for $tge t_0$. Under reasonable conditions on $phi$, we know the solution exists, represented by a curve in the $t$-$y$ plane call this solution $f(t)$. The point $ds (t_0,y_0)$ is of course on this curve. We also know the slope of the curve at this point, namely $dsphi(t_0,y_0)$. If we follow the tangent line for a brief distance, we arrive at a point that should be almost on the graph of $f(t)$, namely $ds(t_0+Delta t, y_0+phi(t_0,y_0)Delta t)$ call this point $ds(t_1,y_1)$. Now we pretend, in effect, that this point really is on the graph of $f(t)$, in which case we again know the slope of the curve through $ds(t_1,y_1)$, namely $dsphi(t_1,y_1)$. So we can compute a new point, $ds(t_2,y_2)=(t_1+Delta t, y_1+phi(t_1,y_1)Delta t)$ that is a little farther along, still close to the graph of $f(t)$ but probably not quite so close as $ds(t_1,y_1)$. We can continue in this way, doing a sequence of straightforward calculations, until we have an approximation $ds(t_n,y_n)$ for whatever time $ds t_n$ we need. At each step we do essentially the same calculation, namely $(t_,y_)=(t_i+Delta t, y_i+phi(t_i,y_i)Delta t).$ We expect that smaller time steps $Delta t$ will give better approximations, but of course it will require more work to compute to a specified time. It is possible to compute a guaranteed upper bound on how far off the approximation might be, that is, how far $ds y_n$ is from $f(t_n)$. Suffice it to say that the bound is not particularly good and that there are other more complicated approximation techniques that do better.

Example 17.4.3 Let us compute an approximation to the solution for $dsdot y=t-y^2$, $y(0)=0$, when $t=1$. We will use $Delta t=0.2$, which is easy to do even by hand, though we should not expect the resulting approximation to be very good. We get eqalign < (t_1,y_1)&=(0+0.2,0+(0-0^2)0.2) = (0.2,0)cr (t_2,y_2)&=(0.2+0.2,0+(0.2-0^2)0.2) = (0.4,0.04)cr (t_3,y_3)&=(0.6,0.04+(0.4-0.04^2)0.2) = (0.6,0.11968)cr (t_4,y_4)&=(0.8,0.11968+(0.6-0.11968^2)0.2) = (0.8,0.23681533952)cr (t_5,y_5)&=(1.0,0.23681533952+(0.6-0.23681533952^2)0.2) = (1.0,0.385599038513605)cr> So $y(1)approx 0.3856$. As it turns out, this is not accurate to even one decimal place. Figure 17.4.1 shows these points connected by line segments (the lower curve) compared to a solution obtained by a much better approximation technique. Note that the shape is approximately correct even though the end points are quite far apart.

If you need to do Euler's method by hand, it is useful to construct a table to keep track of the work, as shown in figure 17.4.2. Each row holds the computation for a single step: the starting point $(t_i,y_i)$ the stepsize $Delta t$ the computed slope $phi(t_i,y_i)$ the change in $y$, $Delta y=phi(t_i,y_i)Delta t$ and the new point, $(t_,y_)=(t_i+Delta t,y_i+Delta y)$. The starting point in each row is the newly computed point from the end of the previous row.

 $(t,y)$ $Delta t$ $phi(t,y)$ $Delta y=phi(t,y)Delta t$ $(t+Delta t,y+Delta y)$ $(0,0)$ .2 (0.2,0)(0.2,0)$.2$ .2$.04$ $(0.4,0.04)$ $(0.4,0.04)$ .2$.3984$ .07968(0.6,0.11968)(0.6,0.11968)$.2$ .58ldots$.117ldots$ $(0.8,0.2368ldots)$ $(0.8,0.236ldots)$ .2$.743ldots$ .148ldots(1.0,0.385ldots$It is easy to write a short function in Sage to do Euler's method. Euler's method is related to another technique that can help in understanding a differential equation in a qualitative way. Euler's method is based on the ability to compute the slope of a solution curve at any point in the plane, simply by computing$phi(t,y)$. If we compute$phi(t,y)$at many points, say in a grid, and plot a small line segment with that slope at the point, we can get an idea of how solution curves must look. Such a plot is called a slope field. A slope field for$dsphi=t-y^2$is shown in figure 17.4.3 compare this to figure 17.4.1. With a little practice, one can sketch reasonably accurate solution curves based on the slope field, in essence doing Euler's method visually. Even when a differential equation can be solved explicitly, the slope field can help in understanding what the solutions look like with various initial conditions. Recall the logistic equation from exercise 13 in section 17.1,$dsdot y = ky(M-y)$:$y$is a population at time$t$,$M$is a measure of how large a population the environment can support, and$k$measures the reproduction rate of the population. Figure 17.4.4 shows a slope field for this equation that is quite informative. It is apparent that if the initial population is smaller than$M$it rises to$M$over the long term, while if the initial population is greater than$M$it decreases to$M\$.