Integrating Factor Method: Finding the Solution to a Cosine Equation

[_N3WTON_]

Homework Statement

Find the general solution to the indicated equation:
[itex] cos(x)y' + ysin(x) = 1 [/itex]

Homework Equations

[itex] e^\int p(x)\,dx * y(x) = int\ f(x)\, dx e^\intp(x)\,dx + C [/itex]

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:
[itex] cos(x)y' + ysin(x) = 1 [/itex]

Homework Equations

[itex] e^\(int p(x))\,dx * y(x) = int\ f(x)\, dxe^\intp(x)\,dx + C [/itex]

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:
[itex] y' + ytan(x) = sec(x) [/itex] and
[itex] p(x) = tan(x) [/itex]
right?

 

[slider142]

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

Yes. The integrating factor of [itex]e^{\int \tan x \,dx}[/itex] 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 [itex]e^{\int \tan x \,dx}[/itex] 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 [itex] e^{sec^2x} * secx [/itex] be integrated? Because that is what I'll end up with no?

 

[slider142]

Edit: I mean can [itex] e^{sec^2x} * secx [/itex] 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 [itex]\frac{\sin x}{\cos x}[/itex].

 

[_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 [itex]\frac{\sin x}{\cos x}[/itex].

ahh...thanks, sorry about the silly mistake

 

[_N3WTON_]

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

 

[slider142]

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

That looks fine. However, it may be preferable to write the solution as [itex]y(x) = \sin x + C\cos x[/itex], 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 [itex]y(x) = \sin x + C\cos x[/itex], 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