Question

Use the method of "undetermined coefficients" to solve the 2nd ODE `y''-2y'=12e^(2x)-8e^(-2x)`?

Answer 1

`y(x) = A + Be^(2x) + 6xe^(2x) - e^(-2x)`

Explanation:

We have:

`y''-2y' = 12e^(2x) - 8e^(-2x)`

This is a second order linear non-Homogeneous Differentiation Equation. The standard approach is to find a solution, `y_c` of the homogeneous equation by looking at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, and then finding an independent particular solution, `y_p` of the non-homogeneous equation.

Complementary Function

The homogeneous equation associated with [A] is

`y''-2y'= 0`

And it's associated Auxiliary equation is:

`m^2-2m = 0`
`m(m-2) = 0`

Which has distinct real root `m=0` and `m=2`

Thus the solution of the homogeneous equation is:

`y_c = Ae^(0x) + Be^(2x)`
`\ \ \ = A + Be^(2x)`

Particular Solution

With this particular equation [A], dur to the fact that `e^(2x)` is already part of the CF then a probable solution is of the form:

`y = axe^(2x) + be^(-2x)`

Where `a,b` are constant coefficients to be determined. Let us assume the above solution works, in which case be differentiating wrt `x` we have:

`y' \ \ = 2axe^(2x)+ae^(2x)-2be^(-2x)`
`\ \ \ \ \ = (2ax+a)e^(2x)-2be^(-2x)`

`y'' = 2(2ax+a)e^(2x) + (2a)e^(2x)+4be^(-2x)`
`\ \ \ \ \ = (4ax+4a)e^(2x) +4be^(-2x)`

Substituting into the initial Differential Equation `[A]` we get:

`{(4ax+4a)e^(2x) +4be^(-2x)} - 2{(2ax+a)e^(2x)-2be^(-2x)} = 12e^(2x) - 8e^(-2x)`

`:. 4axe^(2x)+4ae^(2x) +4be^(-2x) - 4axe^(2x)-2ae^(2x)+4be^(-2x) = 12e^(2x) - 8e^(-2x)`

`:. 4ae^(2x) +4be^(-2x) -2ae^(2x)+4be^(-2x) = 12e^(2x) - 8e^(-2x)`

Equating coefficients of `e^(2x)` and `e^(-2x)` we get:

`e^(2x) \ \ \ : 4a-2a=12 => a=6`
`e^(-2x) : 4b+4b=-8 => b=-1`

And so we form the Particular solution:

`y_p = 6xe^(2x) - e^(-2x)`

General Solution

Which then leads to the GS of [A}

`y(x) = y_c + y_p`
`\ \ \ \ \ \ \ = A + Be^(2x) + 6xe^(2x) - e^(-2x)`