solve d^2y/dx^2+y=tanx

Question

plz help me
aldo i was given a link to read on but i cant even acess it
just help

Answer 1

Really. Ha
you know that the solution of

y" + y = 0

is

y = a sinx + b cosx

Now use the method of variation of parameters:

We now have
y1 = sinx
y2 = cosx

The Wronskian of {sinx,cosx} is -1

u1 = ∫cosx tanx dx = ∫sinx = -cosx
u2 = -∫sinx tanx dx
= ∫cosx - secx dx
= sinx + ln|secx - tanx|

yp = y1 u1 + y2 u2
= sinx (-cosx) + cosx (sinx + ln|secx-tanx|)
= cosx ln|secx-tanx|

So, the general solution is

y = a sinx + b cosx + cosx * ln|secx - tanx|

If this is still confusing, do what I suggested earlier, and google either

y"+y=tanx
or
variation of parameters.
So the general solution is