Differential Equations - Power Series problem with e^t

In summary: In summary, the problem is to solve the differential equation y''+ty'+e^{t}y=0 with initial conditions y(0)=0 and y'(0)=-1. The student's main issue is finding the recursion relation and factoring out t^{whatever} to equate the coefficient to 0. However, the addition of e^{t} complicates this process. Using the Cauchy Product, the student obtains the series ty'=\sum^{\infty}_{k=1}ka_{k}t^{k}, e^{t}=\sum^{\infty}_{k=0}t^{k}/k!, and e^{t}y=\sum^{\infty
  • #1
skyturnred
118
0

Homework Statement



The problem is to solve:

y''+ty'+e[itex]^{t}[/itex]y=0, y(0)=0 and y'(0)=-1

Homework Equations





The Attempt at a Solution



My main issue is the following: I normally find the recursion relation, and then factor out the t[itex]^{whatever}[/itex] and I know that the coefficient to this must equal 0. However, the addition of the e[itex]^{t}[/itex] means I can't factor out ALL of the t[itex]^{whatever}[/itex] so I can't exactly equate the coefficient to 0.

So specifically with the question above, I get the following:

ty'=[itex]\sum[/itex][itex]^{inf}_{k=1}[/itex]ka[itex]_{k}[/itex]t[itex]^{k}[/itex]

e[itex]^{t}[/itex]=[itex]\sum[/itex][itex]^{inf}_{k=0}[/itex]t[itex]^{k}[/itex]/k!

e[itex]^{t}[/itex]y=[itex]\sum[/itex][itex]^{inf}_{k=0}[/itex]t[itex]^{2k}[/itex]a[itex]_{k}[/itex]/k!

y''=[itex]\sum[/itex][itex]^{inf}_{k=2}[/itex]k(k-1)a[itex]_{k}[/itex]t[itex]^{k-2}[/itex]

Then I plug those into the DE, write out the first few terms so that they all start at k=2, and put them all under the same sum to get as follows:

a[itex]_{0}[/itex]+a[itex]_{1}[/itex]t+a[itex]_{1}[/itex]t[itex]^{2}[/itex]+2a[itex]_{2}[/itex]+6a[itex]_{3}[/itex]t+[itex]\sum[/itex][itex]^{inf}_{k=2}[/itex][(k+2)(k+1)a[itex]_{k+2}[/itex]+ka[itex]_{k}[/itex]+a[itex]_{k}[/itex]t[itex]^{k}[/itex]]t[itex]^{k}[/itex]=0

From this you find that a[itex]_{2}[/itex]=-a[itex]_{0}[/itex]/2

and

a[itex]_{1}[/itex]=0

and

a[itex]_{3}[/itex]=0


Like I said, normally at this point I find the recursion relation and factor out a t[itex]^{k}[/itex] so that I know that the entire coefficient is 0. But in this case, after factoring out the t[itex]^{k}[/itex] one of the terms in the coefficient still have a t[itex]^{k}[/itex], so I don't know if I can't just equate this whole thing to 0.

Can anyone help?

Thank-you
 
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  • #2
skyturnred said:

Homework Statement



The problem is to solve:

y''+ty'+e[itex]^{t}[/itex]y=0, y(0)=0 and y'(0)=-1

Homework Equations





The Attempt at a Solution



My main issue is the following: I normally find the recursion relation, and then factor out the t[itex]^{whatever}[/itex] and I know that the coefficient to this must equal 0. However, the addition of the e[itex]^{t}[/itex] means I can't factor out ALL of the t[itex]^{whatever}[/itex] so I can't exactly equate the coefficient to 0.

So specifically with the question above, I get the following:

[itex]ty'=\sum^{\infty}_{k=1}ka_{k}t^{k}[/itex]

You don't need to put the itex tags around every item. Just put them around the whole equation. Right click on the above expression and see how I changed it for you.

e[itex]^{t}[/itex]=[itex]\sum[/itex][itex]^{inf}_{k=0}[/itex]t[itex]^{k}[/itex]/k!

e[itex]^{t}[/itex]y=[itex]\sum[/itex][itex]^{inf}_{k=0}[/itex]t[itex]^{2k}[/itex]a[itex]_{k}[/itex]/k!

You don't multiply two series like that. You need the Cauchy Product. See:
http://planetmath.org/cauchyproduct
 

1. What is a differential equation?

A differential equation is a mathematical equation that relates a function with its derivatives. It is used to describe the relationship between a quantity and its rate of change.

2. What is a power series?

A power series is a series of terms where each term is a constant multiplied by a variable raised to a non-negative integer power. It is commonly used to represent functions as an infinite sum of simpler functions.

3. How do power series relate to differential equations?

Power series can be used to solve differential equations by representing the unknown function as a power series and finding the coefficients that satisfy the given differential equation. This method is known as the method of power series.

4. What is the "e^t" term in the power series?

The "e^t" term represents the exponential function, which is commonly used in many mathematical models and is defined as the sum of the infinite series 1 + t + (t^2)/2! + (t^3)/3! + ..., where t is the variable.

5. How is the "e^t" term used to solve differential equations?

The "e^t" term is often used in the power series method to solve differential equations because it is its own derivative. This means that when the "e^t" term is substituted into a differential equation, it simplifies the equation and makes it easier to solve for the coefficients of the power series.

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