Differential Equations

Question

How do we solve the equation,

(D^2 - 2D + 1)y = e^2x(cosh2x + cos2x)

Answer 1

You first have to express it as (D-1)(D-1)y = e^2x(cosh2x + cos2x) (Call his g(x) )

Then let (D-1)y be expressed as z so that the new expression is (D-1)z = g(x)

Solve for z and then substitute it back to the original expression! This is a tedious process compared to other methods but it does work if the coefficients are constants.

Let's look at a simple example:

Let (D² - 2D +1) = 1

then (D-1)(D-1)y = 1 Let (D-1)y = z, then

(D-1)z = 1 or dz/dx -z = 1 or dz/dx = z+1

The solution to this equation is z = Ke^x - 1 (K is a constant)

Now go back to the original equation and substitute now that we know z

(D-1)y = z = Ke^x - 1 or dy/dx - y = Ke^x - 1

Now you are left with a first order equation that you can solve easily!