Let y=uv, where u=ex, v=sin4(x).
We can write y as y(0)=u(0)v(0), meaning no derivatives of y, u or v.
y(n) represents the nth derivative of y, and similarly for u and v.
sin2(x)=(1-cos(2x))/2, sin4(x)=(1-cos(2x))2/4=(1-2cos(2x)+cos2(2x))/4.
cos2(2x)=(1+cos(4x))/2, so:
v=sin4(x)=¼-½cos(2x)+⅛+⅛cos(4x)=⅜-½cos(2x)+⅛cos(4x).
y(1)=u(0)v(1)+u(1)v(0); y(2)=u(0)v(2)+2u(1)v(1)+u(2)v(0).
y(n)=∑nCru(r)v(n-r) where 0≤r≤n. nCr is the number of combination of r objects out of n objects. This is similar to the formula for binomial expansion of (u+v)n.
u(n)=u=ex.
v(1)=sin(2x)-½sin(4x); v(2)=2cos(2x)-2cos(4x); v(3)=-4sin(2x)+8sin(4x); v(4)=-8cos(2x)+32cos(4x); v(5)=16sin(2x)-128sin(4x)=24sin(2x)-44(½sin(4x)).
v(4n+1)=24nsin(2x)-28n-1sin(4x); v(4n)=-24n-1cos(2x)+28n-3cos(4x);
v(4n-1)=-24n-2sin(2x)+28n-5sin(4x); v(4n-2)=24n-3cos(2x)-28n-7cos(4x); v(4n-3)=24n-4sin(2x)-28n-9sin(4x).
In the following, integer n>0.
y(n)=ex∑nCrv(n-r) for 0≤r≤n. For each v(n-r) we need to use the relevant formula for the derivative based on the cycle of 4.
For example, using n=1, y(3)=ex[v(3)+3v(2)+3v(1)+1]=ex[v(4n-1)+3v(4n-2)+3v(4n-3)+1]=
ex[-24n-2sin(2x)+28n-5sin(4x)+3(24n-3cos(2x)-28n-7cos(4x))+3(24n-4sin(2x)-28n-9sin(4x))+1]=
ex[-4sin(2x)+8sin(4x)+6cos(2x)-6cos(4x)+3sin(2x)-3sin(4x)/2+1]=
ex[-sin(2x)+13sin(4x)/2+6cos(2x)-6cos(4x)+1].