Integrating Factor Method: Finding the Solution to a Cosine Equation

[_N3WTON_]

Homework Statement

Find the general solution to the indicated equation:
$cos(x)y' + ysin(x) = 1$

Homework Equations

$e^\int p(x)\,dx * y(x) = int\ f(x)\, dx e^\intp(x)\,dx + C$

The Attempt at a Solution

Ok, I am having trouble getting started with this problem because I am not really sure what value to assign to p(x). I attempted to assign p(x) = sin(x) therefore u(x) = e^cosx. However, I do not believe that e^cosx can be integrated without using some advanced techniques, so I am quite sure I am not doing the problem correctly.
Edit: when I say e^cosx cannot be integrated I mean not in terms of elementary functions
Edit: Sorry about the bad Latex equations, I am attempting to sort it out (I have never used it before)

 

[LCKurtz]

Homework Statement

Find the general solution to the indicated equation:
$cos(x)y' + ysin(x) = 1$

Homework Equations

$e^\(int p(x))\,dx * y(x) = int\ f(x)\, dxe^\intp(x)\,dx + C$

The Attempt at a Solution

Ok, I am having trouble getting started with this problem because I am not really sure what value to assign to p(x). I attempted to assign p(x) = sin(x) therefore u(x) = e^cosx. However, I do not believe that e^cosx can be integrated without using some advanced techniques, so I am quite sure I am not doing the problem correctly.
Edit: when I say e^cosx cannot be integrated I mean not in terms of elementary functions
Edit: Sorry about the bad Latex equations, I am attempting to sort it out (I have never used it before)

To calculate the integrating factor your equation needs to be in the form$$y' + p(x)y = q(x)$$So you need to start by dividing your equation through by $\cos x$ before trying to evaluate the integrating factor.

 

[_N3WTON_]

To calculate the integrating factor your equation needs to be in the form$$y' + p(x)y = q(x)$$So you need to start by dividing your equation through by $\cos x$ before trying to evaluate the integrating factor.

thank you, so then the equation would become:
$y' + ytan(x) = sec(x)$ and
$p(x) = tan(x)$
right?

 

[slider142]

thank you, so then the equation would become:
$y' + ytan(x) = sec(x)$ and
$p(x) = tan(x)$
right?

Yes. The integrating factor of $e^{\int \tan x \,dx}$ should now work as advertised. :) Just be sure to note whether the points for which cos(x) = 0 are to be included in the domain of the solutions.

 

[_N3WTON_]

Yes. The integrating factor of $e^{\int \tan x \,dx}$ should now work as advertised. :) Just be sure to note whether the points for which cos(x) = 0 are to be included in the domain of the solutions.

Edit: I mean can $e^{sec^2x} * secx$ be integrated? Because that is what I'll end up with no?

 

[slider142]

Edit: I mean can $e^{sec^2x} * secx$ be integrated? Because that is what I'll end up with no?

sec²(x) is not an integral of tan(x). Remember we need to find a function that has tan(x) as its derivative. Try writing tan(x) as $\frac{\sin x}{\cos x}$.

 

[_N3WTON_]

sec²(x) is not an integral of tan(x). Remember we need to find a function that has tan(x) as its derivative. Try writing tan(x) as $\frac{\sin x}{\cos x}$.

ahh...thanks, sorry about the silly mistake

 

[_N3WTON_]

ok, for my final answer I am getting:
$y(x) = sin(x) + \frac {C}{sec(x)}$

 

[slider142]

ok, for my final answer I am getting:
$y(x) = sin(x) + \frac {C}{sec(x)}$

That looks fine. However, it may be preferable to write the solution as $y(x) = \sin x + C\cos x$, since both this function and the original differential equation is defined for all x, while your function is not defined for some of the values of x for which the original differential equation is defined. :)

 

[_N3WTON_]

That looks fine. However, it may be preferable to write the solution as $y(x) = \sin x + C\cos x$, since both this function and the original differential equation is defined for all x, while your function is not defined for some of the values of x for which the original differential equation is defined. :)

thanks again, I'll do that