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# Compute the coefficients of a given power series

Consider the power series

with coefficients determined by the identity

Compute the coefficients and determine the sum of the series.

We know the power-series expansion for is given by

Starting with and the given identity we can compute the coefficients by equating the coefficients of like powers of ,

Then from the identity for the coefficients (and noting that the series converges absolutely for all real so we may split the sum into separate sums without worry),

This is a first order linear differential equation of the form . Furthermore, the initial condition implies that when . Therefore, the solution is

where

So, we have

# Assume y′′ + xy′ + y = 0 has a power-series solution and determine the coefficient an

Assume that the differential equation

has a power-series solution and find a formula for the coefficient .

First, we have

So, from the given differential equation we have

Since each coefficient of must equal 0 for this equation to hold we have

By induction we then have

The coefficients and are arbitrary and we denote them by and respectively. Then we have

# Assume y′′ = xy has a power-series solution and determine the coefficient an

Assume that the differential equation

has a power-series solution and find a formula for the coefficient .

First, we have

Therefore, we have

Equating like powers of , we have a recursive relation for when given by

Furthermore, we have that and that and are arbitrary constants, say and , respectively. Then by induction we establish

Therefore,

# Assume y′ = α y has a power series solution and determine the coefficient an

Assume that the differential equation

has a power-series solution and find a formula for the coefficient .

First, we have

Therefore, we have

Equating like powers of we obtain the recursive relation

By induction, we then have

Therefore,

# Find the first four nonzero terms of the power series solution of y′ = x + y2

Consider the differential equation

with initial conditions when . Assume this differential equation has a power-series solution and compute the first four nonzero terms of the expansion.

Let

be the power-series solution of the differential equation. Then we must have

From the initial conditions we know . Then, equating like powers of we can solve for the first four nonzero terms in the power series expansion:

(Note: I think the solution in the back of Apostol is wrong on this. Apostol has , , and . I’m going to mark this as errata until someone convinces me Apostol is actually correct.)

Therefore, we have

# Find the first four nonzero terms of the power series solution of y′ = 1 + xy2

Consider the differential equation

with initial conditions when . Assume this differential equation has a power-series solution and compute the first four nonzero terms of the expansion.

Let

be the power series solution of the differential equation. Then we must have

From the initial condition when we know . Therefore, equating like powers of we have

(Note: The book gives the value . I think the answer we have above is correct. I’m marking this as errata unless someone convinces me that Apostol is correct.)

Therefore, we have

# Find the first four nonzero terms of the power series solution of y′ = x2 + y2

Consider the differential equation

with initial conditions when . Assume this differential equation has a power-series solution and compute the first four nonzero terms of the expansion.

Let

be the power-series solution of the differential equation. Then we must have

From the initial condition when we have . Therefore, equating like powers of we have the following equations

Therefore, we have

# Prove that the Bessel functions are solutions of the Bessel equation

In the previous exercise (Section 11.16, Exercise #10) we defined the Bessel functions of the first kind of orders zero and one by,

Prove that these Bessel functions are solutions of the differential equation

when and , respectively.

Proof. In the previous exercise (linked above) we proved the following

For the case we have the differential equation

Plugging in we then have

So, is indeed a solution in the case .

Now, from the previous exercise we have the relations

Starting with the case we differentiate,

Using the relations above from the previous problem, we substitute

Hence, is indeed a solution of the differentiation equation

# Determine the interval of convergence of a given power series and show that it satisfies a differential equation

Consider the function defined by the power series

Determine the interval of convergence for the power series and show that satisfies the differential equation

First, we use the ratio test to determine the interval of convergence,

Hence, the series converges for all . Then, to show that it satisfies the given differential equation we take the first two derivatives,

Then we have

Therefore, indeed is a solution of the given differential equation.

Now, to find the sum we first need to get the general form of the solutions for the differential equation

First, we find the general form of the solutions of the homogeneous equation

In this case we have an equation of the form where and . From this we can compute and . Therefore, the general form of the solutions is

Then, we can find a particular solution of the given equation by inspection since

is a solution. Therefore, the general solution to the given inhomogeneous equation is

Now, in the particular case we also have the initial condition and so we have

Furthermore, since is an odd function we must have

Therefore, we conclude and . And so,

# Determine the interval of convergence for a given power series and show that it satisfies a given differential equation

Consider the function defined by the power series

Determine the interval of convergence for and show that it satisfies the differential equation

First, to determine the interval of convergence for the power series we use the ratio test

Hence, the series converges for all . Next, to show that it satisfies the given differential equation we take the first two derivatives,

Then, we have

Thus, indeed satisfies the given differential equation.

Further, in a previous exercise (Section 8.14, Exercise #2) that the solution of the differential equation are all of the form

For this problem we also have so

Finally, we know this function is an even function (since for all because we have inside the sum is the only term). This means we must have . Hence, we must have