does anyone know how to solve this initial value problem? −8y′′−8y=sin(x);y(0)=1,y′(0)=−3?
−8y′′−8y=sin(x);y(0)=1,y′(0)=−3 it seems when I solve for the general solution I get 0
Update:I mean the particular solution
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I'm not incredibly sure what you are necessarily asking, but I'm fairly confident that I can solve for y" as -1. So, I'll describe how I did that, maybe it helps.
-8y" - 8y = sin(x) so, I kind of just plugged and chugged
-8y" - 8(1) = sin(0)
simplifying:
-8y" - 8 = 0
again:
-8y" = 8
y" = -1
therefore y"(0)=-1
Hope that helps, somewhat.