Question

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

Answer 1

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`