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