$ 1/x = ln x (true)
but for $ 1/tanx you would need the chain rule - in fact, think of it as $ cos x / sin x...
Just in case a picture helps...
Spoiler:
... do you have the IF?
Key:Spoiler:
Hi guys, this is the differentioan equation I need help to solve
y' + y cotx = sinx
attempt:¨
'
This is the case of using an Integration factor as far as I am concerned.
The form of the equation is
dy/dx (with constant 1) + a(x) * y = b(x)
1. I can see that a(x) is in fact cot x . my integration factor will be e^$cot x dx
as cot x is the same as 1/tanx I put it in the I.F instead
Thus e^$1/tanx * dx -> e^ln tan x = I.F = tanx
2. Then i multiply it
tanx * y' + tanx * y cot x = tanx * sinx
as far as I can see, this is not the product rule?
Can someone help me solve this?
$ 1/x = ln x (true)
but for $ 1/tanx you would need the chain rule - in fact, think of it as $ cos x / sin x...
Just in case a picture helps...
Spoiler:
... do you have the IF?
Key:Spoiler:
so sinx is the I.F --->
sinx * y' + sinx * y* cosx/sinx = sinx^2
sinx * y' + y*cosx = sinx^2
(sinx * y)' = (sinx)^2
After integrating
(sinx * y) = 1/2 * (x-sinx*cosx)
y = 1/2 * (x-sinx*cosx)
--------------------
sinx
[B]Wrong or right? could you possibly solve it for me?[B]