Asked by dams
solve d^2y/dx^2+y=tanx
plz help me
aldo i was given a link to read on but i cant even acess it
just help
plz help me
aldo i was given a link to read on but i cant even acess it
just help
Answers
Answered by
Steve
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
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
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