What is a solution to the differential equation #y'' + 4y = 8sin2t#?

1 Answer
Nov 1, 2016

The General Solution to the DE # y''+4y=8sin2t # is
# y = Ccos2t+Dsin2t -2tcos2t#

Explanation:

There are two major steps to solving Second Order DE's of this form:
# y''+4y=8sin2t #

Find the Complementary Function (CF)
This means find the general solution of the Homogeneous Equation
# y''+4y = 0 #

To do this we look at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, i.e.

As # y''+0y' +4y = 0 #, the Axillary Equation is:
#m^2+0m+4=0#
#m^2+4=0#

This has two distinct complex solutions, #m=0+-2i#

And so the solution to the DE is;
# y = e^(0t){ Ccos2t+Dsin2t} # Where #C,D# are arbitrary constants
# y = Ccos2t+Dsin2t #

-+-+-+-+-+-+-+-+-+-+-+-+-+-+
Verification:
If # y = Ccos2t+Dsin2t #, then
# y' = -2Csin2t+2Dcos2t #
# y'' = -4Ccos2t+-4Dsin2t #

And so, # y''+4y = -4Ccos2t-4Dsin2t + 4(Ccos2t+Dsin2t) = 0 #
-+-+-+-+-+-+-+-+-+-+-+-+-+-+

Find a Particular Integral* (PI)

This means we need to find a specific solution (that is not already part of the solution to the Homogeneous Equation).

We would normally look for a particular solution which is a combination of functions on the RHS of the form
# y=Asin2t+Bcos2t #

However these are both solutions of the homogeneous system, so instead we must look for solution of the form
# y=Atsin2t+Btcos2t #

If # y=Atsin2t+Btcos2t #, then (using the product rule)
# y'=(At)(2cos2t) + (sin2t)(A) + (Bt)(-2sin2t) + (cos2t)(B) #
# :. y'=2Atcos2t + Asin2t - 2Btsin2t + Bcos2t #

Differentiating again, we get:
# y''= (2At)(-2sin2t) + (cos2t)(2A) + 2Acos2t - { (2Bt)(2cos2t) + (sin2t)(2B) } -2Bsin2t #

# :. y''= -4Atsin2t + 2Acos2t + 2Acos2t - 4Btcos2t) -2Bsin2t -2Bsin2t #
# :. y''= -4Atsin2t + 4Acos2t 4Btcos2t) -4Bsin2t #

If we substitute into the DE # y''+4y=8sin2t # we get:
# -4Atsin2t + 4Acos2t - 4Btcos2t -4Bsin2t + 4{ Atsin2t+Btcos2t } = 8sin2t #

# :. -4Atsin2t + 4Acos2t - 4Btcos2t -4Bsin2t + 4Atsin2t+4Btcos2t = 8sin2t #

# :. 4Acos2t - 4Bsin2t = 8sin2t #

Equating Coefficients of #cos2t# and #sin2t# we have
# 4A=0 => A=0 #
# -4B=8 => B=-2 #

So we have found that a Particular Solution is:
# y=Atsin2t+Btcos2t #
ie # y=-2tcos2t #

General Solution (GS)
The General Solution to the DE is then:
GS = CF + PI

Hence The General Solution to the DE # y''+4y=8sin2t # is
# y = Ccos2t+Dsin2t -2tcos2t#