Helping students gain understanding and self-confidence in algebra.
doggy wrote:I'm having a bit of trouble figuring out this differential equation. The question is Find the particular solution of the differential equation:
xy ' +4y = 15xcos(x^5) sastisfying the initial condition y(pi)=0.
doggy wrote:I have gotten up to that point as well but I'm having trouble with the conditions when I sub in pi I got:
3 cos(pi)^5/(pi)^4 + c/(pi/4) =0
thus c= -3 cos((pi)^5)
So the final answer should be :
3 cos (x^5) - 3 cos(pi)^5 but that's not correct
doggy wrote:Don't I need the cos (pi)^5?
now take care of the initial condition
doggy wrote:So then the final answer would be c=0
and y= 3 sin(x^5)/x^4?