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Use power series to solve y = 2 y , y ( 0 ) = 5 .

y = 5 e 2 x

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We now consider an example involving a differential equation that we cannot solve using previously discussed methods. This differential equation

y x y = 0

is known as Airy’s equation . It has many applications in mathematical physics, such as modeling the diffraction of light. Here we show how to solve it using power series.

Power series solution of airy’s equation

Use power series to solve

y x y = 0

with the initial conditions y ( 0 ) = a and y ( 0 ) = b .

We look for a solution of the form

y = n = 0 c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + c 4 x 4 + .

Differentiating this function term by term, we obtain

y = c 1 + 2 c 2 x + 3 c 3 x 2 + 4 c 4 x 3 + , y = 2 · 1 c 2 + 3 · 2 c 3 x + 4 · 3 c 4 x 2 + .

If y satisfies the equation y = x y , then

2 · 1 c 2 + 3 · 2 c 3 x + 4 · 3 c 4 x 2 + = x ( c 0 + c 1 x + c 2 x 2 + c 3 x 3 + ) .

Using [link] on the uniqueness of power series representations, we know that coefficients of the same degree must be equal. Therefore,

2 · 1 c 2 = 0 , 3 · 2 c 3 = c 0 , 4 · 3 c 4 = c 1 , 5 · 4 c 5 = c 2 , .

More generally, for n 3 , we have n · ( n 1 ) c n = c n 3 . In fact, all coefficients can be written in terms of c 0 and c 1 . To see this, first note that c 2 = 0 . Then

c 3 = c 0 3 · 2 , c 4 = c 1 4 · 3 .

For c 5 , c 6 , c 7 , we see that

c 5 = c 2 5 · 4 = 0 , c 6 = c 3 6 · 5 = c 0 6 · 5 · 3 · 2 , c 7 = c 4 7 · 6 = c 1 7 · 6 · 4 · 3 .

Therefore, the series solution of the differential equation is given by

y = c 0 + c 1 x + 0 · x 2 + c 0 3 · 2 x 3 + c 1 4 · 3 x 4 + 0 · x 5 + c 0 6 · 5 · 3 · 2 x 6 + c 1 7 · 6 · 4 · 3 x 7 + .

The initial condition y ( 0 ) = a implies c 0 = a . Differentiating this series term by term and using the fact that y ( 0 ) = b , we conclude that c 1 = b . Therefore, the solution of this initial-value problem is

y = a ( 1 + x 3 3 · 2 + x 6 6 · 5 · 3 · 2 + ) + b ( x + x 4 4 · 3 + x 7 7 · 6 · 4 · 3 + ) .
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Use power series to solve y + x 2 y = 0 with the initial condition y ( 0 ) = a and y ( 0 ) = b .

y = a ( 1 x 4 3 · 4 + x 8 3 · 4 · 7 · 8 ) + b ( x x 5 4 · 5 + x 9 4 · 5 · 8 · 9 )

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Evaluating nonelementary integrals

Solving differential equations is one common application of power series. We now turn to a second application. We show how power series can be used to evaluate integrals involving functions whose antiderivatives cannot be expressed using elementary functions.

One integral that arises often in applications in probability theory is e x 2 d x . Unfortunately, the antiderivative of the integrand e x 2 is not an elementary function. By elementary function, we mean a function that can be written using a finite number of algebraic combinations or compositions of exponential, logarithmic, trigonometric, or power functions. We remark that the term “elementary function” is not synonymous with noncomplicated function. For example, the function f ( x ) = x 2 3 x + e x 3 sin ( 5 x + 4 ) is an elementary function, although not a particularly simple-looking function. Any integral of the form f ( x ) d x where the antiderivative of f cannot be written as an elementary function is considered a nonelementary integral    .

Nonelementary integrals cannot be evaluated using the basic integration techniques discussed earlier. One way to evaluate such integrals is by expressing the integrand as a power series and integrating term by term. We demonstrate this technique by considering e x 2 d x .

Using taylor series to evaluate a definite integral

  1. Express e x 2 d x as an infinite series.
  2. Evaluate 0 1 e x 2 d x to within an error of 0.01 .
  1. The Maclaurin series for e x 2 is given by
    e x 2 = n = 0 ( x 2 ) n n ! = 1 x 2 + x 4 2 ! x 6 3 ! + + ( −1 ) n x 2 n n ! + = n = 0 ( −1 ) n x 2 n n ! .

    Therefore,
    e x 2 d x = ( 1 x 2 + x 4 2 ! x 6 3 ! + + ( −1 ) n x 2 n n ! + ) d x = C + x x 3 3 + x 5 5.2 ! x 7 7.3 ! + + ( −1 ) n x 2 n + 1 ( 2 n + 1 ) n ! + .
  2. Using the result from part a. we have
    0 1 e x 2 d x = 1 1 3 + 1 10 1 42 + 1 216 .

    The sum of the first four terms is approximately 0.74 . By the alternating series test, this estimate is accurate to within an error of less than 1 216 0.0046296 < 0.01 .
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Questions & Answers

what is math number
Tric Reply
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Sidiki Reply
Need help solving this problem (2/7)^-2
Simone Reply
x+2y-z=7
Sidiki
what is the coefficient of -4×
Mehri Reply
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
Alfred Reply
An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
Kala Reply
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
Moses Reply
12, 17, 22.... 25th term
Alexandra Reply
12, 17, 22.... 25th term
Akash
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Carole
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AJ
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Atone
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Adu
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Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
kinnecy Reply
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
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is it a question of log
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Commplementary angles
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Nharnhar
Leaves accumulate on the forest floor at a rate of 2 g/cm2/yr and also decompose at a rate of 90% per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor, assuming at time 0 there is no leaf litter on the ground. Does this amount approach a steady value? What is that value?
Abdul Reply
You have a cup of coffee at temperature 70°C, which you let cool 10 minutes before you pour in the same amount of milk at 1°C as in the preceding problem. How does the temperature compare to the previous cup after 10 minutes?
Abdul
Practice Key Terms 2

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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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