Differential Equation - Nonhomogenous Hermite's Polynomial

In summary, to find the general solution up to degree 6 of the given equation, you can use the power series method and equate the series for y'', ty', and y on one side to the series for e-2t on the other side. Start with the assumption that y = a0 + a1t + a2t2 + ... up to degree 8, then calculate y' and y'', multiply y' by t, and group them by like powers of t. On the right side, the power series for e-2t will have terms up to degree 6. Equate both sides and solve for the coefficients to get the general solution. Don't forget to include the degree 0 term on
  • #1
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Homework Statement



Find the general solution up to degree 6 of y'' + ty' + y = e-2t

Homework Equations





The Attempt at a Solution



I know how to solve it for y'' + ty' + y = 0, but what do I do about the e-2t?
 
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  • #2
Are you doing the homogeneous equation by a power series method? If so, combine your series for y'', ty', and y on the one side and equate them to the series for e-2t.
 
  • #3
am I supposed to use taylor's polynomials for the e-2t after?
 
  • #4
That's what I would do for starters. After all, you're only concerned with terms up to degree 6.

Start with the assumption that y = a0 + a1t + a2t2 and so on up to degree 8 or so (to get degree 6 term in your second derivative. Calculate y' and y'' and multiply y' by t, then add them together and group them by like powers of t. On the right side, you'll have the power series for e-2t, which looks like 1 - 2t + (2t)2/2! + a few more terms to get to degree 6. Then equate both sides and solve for your coefficients.
 
  • #5
So would each group of t be equal to degree 6 of e-2t power series.

For example

(...)t = degree 6 of e-2t
(...)t2 = degree 6 of e-2t
(...)t3 = degree 6 of e-2t

and so forth

or would it look like this:

(...)t = degree 1 of e-2t
(...)t2 = degree 2 of e-2t
(...)t3 = degree 3 of e-2t

and so forth
 
  • #6
This one.
(...)t = degree 1 of e-2t
(...)t2 = degree 2 of e-2t
(...)t3 = degree 3 of e-2t
and so forth
Don't forget the degree 0 term on each side.
 

1. What is a nonhomogenous Hermite's polynomial differential equation?

A nonhomogenous Hermite's polynomial differential equation is a type of differential equation that involves a polynomial function multiplied by a nonhomogenous term. It is typically in the form of y'' + p(x)y' + q(x)y = r(x), where r(x) is the nonhomogenous term.

2. What is the difference between a homogenous and nonhomogenous Hermite's polynomial differential equation?

A homogenous Hermite's polynomial differential equation only has a polynomial function in the equation, while a nonhomogenous one also includes a nonhomogenous term. This nonhomogenous term can be a constant or a function of x.

3. How do you solve a nonhomogenous Hermite's polynomial differential equation?

To solve a nonhomogenous Hermite's polynomial differential equation, you can use the method of undetermined coefficients or variation of parameters. In the method of undetermined coefficients, you make an educated guess for the particular solution based on the nonhomogenous term, while in variation of parameters, you find a complementary function and a particular solution separately.

4. What is the significance of Hermite's polynomials in solving differential equations?

Hermite's polynomials are a set of orthogonal polynomials that have many applications in physics and engineering, including solving differential equations. They can be used as a basis for solving nonhomogenous Hermite's polynomial differential equations.

5. Can a nonhomogenous Hermite's polynomial differential equation have multiple solutions?

Yes, a nonhomogenous Hermite's polynomial differential equation can have multiple solutions. This is because the particular solution obtained using the method of undetermined coefficients may differ depending on the nonhomogenous term, and the complementary function can also have different forms depending on the roots of the characteristic equation.

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