How do you solve #y''+3y'-4y=100cos(2x)#?

1 Answer
Steve M
Jul 2, 2017

# y=Ae^x+Be^(-4x)-8cos(2x)+6sin(2x)#

Explanation:

We have:

# y'' + 3y'-4y=100cos(2x)# ..... [A]

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.

Complimentary Function

The homogeneous equation associated with [A] is

# y'' + 3y'-4y = 0 #

And it's associated Auxiliary equation is:

# m^2+3m-4 = 0 #

Which has two real and distinct solutions #m=-4,1#

Thus the solution of the homogeneous equation is:

# y_c = Ae^x+Be^(-4x) #

Particular Solution

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

# y = acos(2x)+bsin(2x) #

Where #a# and #b# are constants to be determined by substitution

Let s assume the above solution works, in which case be differentiating wrt #x# we have:

# y' \ \= -2asin(2x)+2bcos(2x) #
# y'' = -4acos(2x)-4bsin(2x) #

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

# -4acos(2x)-4bsin(2x) + 3{-2asin(2x)+2bcos(2x)} - 4{acos(2x)+bsin(2x)}=100cos(2x)#

# :. -4acos(2x)-4bsin(2x) -6asin(2x)+6bcos(2x) - 4acos(2x)-4bsin(2x)=100cos(2x)#

Equating coefficients of #cos(2x)# and #sin(2x)# we get:

#cos(2x): -4a+6b-4a=100 => -8a+6b=100#
#sin(2x): -4b -6a-4b = 0 \ \ \ \ \ => -6a-8b=0 #

Solving simultaneously we get:

# a=-8 # and #b=6#

And so we form the Particular solution:

# y_p = -8cos(2x)+6sin(2x)#

Which then leads to the GS of [A}

# y(x) = y_c + y_p #
# \ \ \ \ \ \ \ = Ae^x+Be^(-4x)-8cos(2x)+6sin(2x)#

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