Question

Use the method of "undetermined coefficients" to solve the 2nd ODE `y''+4y'+4y=e^(-2x)sin2x` ?

Answer 1

`y(x) = Axe^(-2x)+Be^(-2x) -1/4e^(-2x)sin2x`

Explanation:

We have:

`y''+4y'+4y = e^(-2x)sin2x`

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''+4y'+4y`

And it's associated Auxiliary equation is:

`m^2+4m+4 = 0`
`(m+2)^2 = 0`

Which has a repeated real root `m=-2`

Thus the solution of the homogeneous equation is:

`y_c = (Ax+B)e^(-2x)`

Particular Solution

With this particular equation [A], a probable solution is of the form:

`y = e^(-2x)(acos2x + bsin2x)`

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' \ \ = e^(-2x)(-2asin2x + 2bcos2x) -2e^(-2x)(acos2x + bsin2x)`
`\ \ \ \ \ = e^(-2x)((2b-2a)cos2x - (2a+2b)sin2x)`

And:

`y'' = e^(-2x)(-2(2b-2a)sin2x - 2(2a+2b)cos2x) -2 e^(-2x)((2b-2a)cos2x - (2a+2b)sin2x)`

`\ \ \ \ \ = e^(-2x)(-8b)cos2x + (8a)sin2x)`

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

`{e^(-2x)(-8b)cos2x + (8a)sin2x)} + 4{e^(-2x)((2b-2a)cos2x - (2a+2b)sin2x)} + 4{e^(-2x)(acos2x + bsin2x)} = e^(-2x)sin2x`

Equating coefficients of `cos2x` and `sin2x` we get:

`cos2x: -8b + 8b-8a + 4a =0`
`sin2x: 8a - 8a-8b+4b=1`

Solving simultaneous we have:

`a = 0, b=-1/4`

And so we form the Particular solution:

`y_p = -1/4e^(-2x)sin2x`

General Solution

Which then leads to the GS of [A}

`y(x) = y_c + y_p`
`\ \ \ \ \ \ \ = (Ax+B)e^(-2x) -1/4e^(-2x)sin2x`
`\ \ \ \ \ \ \ = Axe^(-2x)+Be^(-2x) -1/4e^(-2x)sin2x`