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

Nov 1, 2016

The General Solution to the DE $y ' ' + 4 y = 8 \sin 2 t$ is
$y = C \cos 2 t + D \sin 2 t - 2 t \cos 2 t$

Explanation:

There are two major steps to solving Second Order DE's of this form:
$y ' ' + 4 y = 8 \sin 2 t$

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

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

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

This has two distinct complex solutions, $m = 0 \pm 2 i$

And so the solution to the DE is;
$y = {e}^{0 t} \left\{C \cos 2 t + D \sin 2 t\right\}$ Where $C , D$ are arbitrary constants
$y = C \cos 2 t + D \sin 2 t$

-+-+-+-+-+-+-+-+-+-+-+-+-+-+
Verification:
If $y = C \cos 2 t + D \sin 2 t$, then
$y ' = - 2 C \sin 2 t + 2 D \cos 2 t$
$y ' ' = - 4 C \cos 2 t \pm 4 D \sin 2 t$

And so, $y ' ' + 4 y = - 4 C \cos 2 t - 4 D \sin 2 t + 4 \left(C \cos 2 t + D \sin 2 t\right) = 0$
-+-+-+-+-+-+-+-+-+-+-+-+-+-+

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
$y = A \sin 2 t + B \cos 2 t$

However these are both solutions of the homogeneous system, so instead we must look for solution of the form
$y = A t \sin 2 t + B t \cos 2 t$

If $y = A t \sin 2 t + B t \cos 2 t$, then (using the product rule)
$y ' = \left(A t\right) \left(2 \cos 2 t\right) + \left(\sin 2 t\right) \left(A\right) + \left(B t\right) \left(- 2 \sin 2 t\right) + \left(\cos 2 t\right) \left(B\right)$
$\therefore y ' = 2 A t \cos 2 t + A \sin 2 t - 2 B t \sin 2 t + B \cos 2 t$

Differentiating again, we get:
$y ' ' = \left(2 A t\right) \left(- 2 \sin 2 t\right) + \left(\cos 2 t\right) \left(2 A\right) + 2 A \cos 2 t - \left\{\left(2 B t\right) \left(2 \cos 2 t\right) + \left(\sin 2 t\right) \left(2 B\right)\right\} - 2 B \sin 2 t$

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

If we substitute into the DE $y ' ' + 4 y = 8 \sin 2 t$ we get:
$- 4 A t \sin 2 t + 4 A \cos 2 t - 4 B t \cos 2 t - 4 B \sin 2 t + 4 \left\{A t \sin 2 t + B t \cos 2 t\right\} = 8 \sin 2 t$

$\therefore - 4 A t \sin 2 t + 4 A \cos 2 t - 4 B t \cos 2 t - 4 B \sin 2 t + 4 A t \sin 2 t + 4 B t \cos 2 t = 8 \sin 2 t$

$\therefore 4 A \cos 2 t - 4 B \sin 2 t = 8 \sin 2 t$

Equating Coefficients of $\cos 2 t$ and $\sin 2 t$ we have
$4 A = 0 \implies A = 0$
$- 4 B = 8 \implies B = - 2$

So we have found that a Particular Solution is:
$y = A t \sin 2 t + B t \cos 2 t$
ie $y = - 2 t \cos 2 t$

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

Hence The General Solution to the DE $y ' ' + 4 y = 8 \sin 2 t$ is
$y = C \cos 2 t + D \sin 2 t - 2 t \cos 2 t$