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

1 Answer
Dec 28, 2017

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