Is e^x the Optimal Integrating Factor for Solving Differential Equations?

[asdf1]

for the question, siny+cosydy=0, i want to find an integrating factor.
my work:
(1/F)(dF/dx)=(1/cosy)(cosy+siny)=1+tany
=>lny=x +xtany +c`
=> y =ce^(x+xtany)
however, the question wants the integrating factor to be e^x...
why?

 

[HallsofIvy]

Can I assume you mean sinydx+ cosydy= 0? Without a dx in there, it doesn't make sense. If that's the case, then an obvious integrating factor is 1/siny since multiplying through by that gives dx+ (cosx/sinx)dy= 0 which is clearly exact.

I don't know what you mean by "the question wants the integrating factor to be e^x"!
I wasn't aware that questions wanted anything!

 

[Corneo]

Isn't that equation seperable?

 

[asdf1]

opps! I'm sorry for the mistype! :P
you're right, it's "sinydx+ cosydy= 0"
that question wanted to prove that the integrating factor is e^x, but the integrating factor that i found was y =ce^(x+xtany)...

 

[lurflurf]

so you want an integrating factor u such that
$$\frac{\partial}{\partial y}u\sin(y)=\frac{\partial}{\partial x}u\cos(y)$$
or
$$\frac{\partial u}{\partial y}\sin(y)+u\cos(y)=\frac{\partial u}{\partial x}\cos(y)$$
integrating factors are not unique so assume
$$\frac{\partial u}{\partial y}=0$$

 

[HallsofIvy]

If the problem says "show that e^x is an integrating factor", thenyou don't have to find the integrating factor yourself (as lurflurf said, integrating factors are not unique), just multiply the equation by e^x and see if the result is exact.

If you got ce^(x+xtany) as an integrating factor, you sure like doing things the hard way! As I said earlier, 1/sin y is an obvious integrating factor (because, as Corneo said, the equation is separable. Multiplying by
1/sin y "separates" it)

 

[asdf1]

lol...
i didn't think of that...
thank you very much! :)