Dif.eq. with trigonometric functions involving y

[mausmust]

I tried to solve it but confused. Pls. help me to solve this equation:

dy/dx + (e^x)*Sec(y) = Tan(x);

(hint: integrating factor is e^-ax, and a is unknown, a ε ℝ, find it, solve the equation)

Thnx.

 

[tiny-tim]

welcome to pf!

hi mausmust! welcome to pf!

(try using the X² button just above the Reply box )

show us your full calculations, and then we'll see what went wrong, and we'll know how to help!

 

[mausmust]

Hii, thank you :)

I did:

dy + (e^xSec(y)-Tan(x))dx = 0

0 ≠ e^xSec(y)Tan(y) (∂N/∂x ≠ ∂M/∂y) So, we need an integrating factor. If we use e^-ax;

-ae^-ax = e^(1-ax)Sec(y)Tan(y) appears.

Shouldn't be "a" is a real value? Also;

I tried to modify equation with ArcSecant function to become a linear equation with this form;

dy/dx + P(x)y = Q(x)

but it went more complicated. How can we solve it?

 

[jackmell]

I don't think you're doin' that right. Also, don't use capitals letters for the trig functions. If you have:

$$Mdx+Ndy=0$$

and:

$$\frac{1}{N}\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)=f(x)$$

then the integrating factor is:

$$u=\exp\left(\int f(x)dx\right)$$

but that's not just $e^{-ax}$ for real a.