Question

du(x)/dx- tan(x)u(x)=-2sinx

find u(x)

inhomogeneous differential equation

Answer 1

lol wtf

Answer 2

actually im trying to solve inhomogeneous differential equation

Answer 3

du(x)/dx- tan(x)u(x)=-2sinx

Answer 4

Is that calculus or..?

Answer 5

advance applied mathematics

Answer 6

I think the integral you looking for is:
\[e^{\int\limits\tan x}\]
which would be integrating factor for this equation

Answer 7

well the answer appears to be u(x)=c(1/cos(x))+cos(x)

Answer 8

but i dont know how to get to there

Answer 9

I am a bit rusty on this, but here it goes (as much as I remmeber)
1º Find the integrating factor of the 1º order linear nonhomogeneous differential equation u'-Pu=Q, which is e^{int P}. Inthis case it is equal 1/cos x.
2º Multiply bouth side of the equation by this factor. u'/cosx -u(tanx/cosx)=-2sinx/cosx
3º Notice that the left side is derivative of u/cosx, so d(u/cosx)/dx = -2tanx
4º Integrate bouth sides: u/cosx=-2 int tanx

Answer 10

I think I made a mistake in integrating factor is not 1/cosx, but just cosx
and the answer should be u=2+c/cosx I think

Answer 11

just tried but it dint give the desired answer...

Answer 12

u(x)=c(1/cos(x))+cos(x)
this was the given answer

Answer 13

Integrating factor = cosx
after multiply:
cosx u' -sinx u =-2sinx cosx
d(cosx u)/dx = - 2sinx cosx
cosx u =-2 int {sinx cosx dx }
cosx u = 2 cos x + C
u=2 +C/cosx

Answer 14

yeah im getting the same answer as yours..im wondering why does the book is written as u=cosx +C/cosx

Answer 15

maybe I made a mistake somewhere............

Answer 16

hmm not so sure too...but thanks alot @myko