a question on differential equations. finding a general solution to dy/dx +2ytanx = sinx.?

Question

i have found the integrating factor a(x) = exp(int(2tanx)dx)=sec^2 x.
but the integral i get i realy don't have an idea how to solve it, it may be simple but i don't see it or i've done it wrong as a whole please could someone help. thank you.

Answer 1

dy/dx +2ytanx = sinx

the easiest way to do this
problem is the integrating
factor method
dy/dx+g(x)y=h(x)
1)find the integrating factor p
p=e^(int2tanxdx)
=e^(-2ln(cosx)=e^(2ln(secx))
=(secx)^2
2)rewrite the DE as
d/dx(secx)^2*y=(secx)^2*sinx
3)integrate the last equation to
obtain
(secx)^2*y
=int((secx)^2*sinx)dx
=int(sinx/(cosx)^2)dx
=int(tanx*secx)dx
=secx+C
4)divide through by (secx)^2,to
obtain the general solution in
explicit form
therefore,
y=(secx+C)/(secx)^2
=cosx+C*(cosx)^2
=cosx(1+C*cosx)

i hope that you can follow my
working
substitute the value for y into
the original equation to make
sure that the solution is correct

{dy/dx= -sinx-2Ccosxsinx
2ytanx
=2sinx+2Csinxcosx
therefore,
dy/dx+2ytanx=sinx
as required}

Answer 2

dy/dx + (2tanx) y = sinx

Let P = 2 tanx = 2 sinx/cosx

∫Pdx = - 2 log(cosx) = log(1/cos²x)

Integrating Factor = e^log(1/cos²x) = 1/cos²x

Multiply both sides by Integrating Factor and integrate: -

y = ∫ (sin x/cos²x) dx = ∫ tan x sec x dx = sec x + c

Answer 3

Implicit Integration? (dunno the name)

try to get the y terms on one side with the dy/dx and the x on the other sides.

dy/dx = sin x - 2ysin x/cos x

dy/dx = sin x ([cos x-2y]/cos x)

(1-2y)/cos x .dy = sin x. dx

(1-2y).dy = sin x cos x . dx

But sin x cos x = 0.5sin[2x]

(1-2y).dy = 0.5sin2x

integrate both sides (ignoring +C on LHS for now)

y - y^2 = 0.25cos[2x] + C

Thats the integral I get. Where C is the constant of Integration

damn. I got it wrong =[
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