What is the quickest way to solve this 2nd Order Linear ODE?

This appeared on my professor's test review, and its taken me hours to, surprise surprise, get the wrong answer. Could someone help me with the method I should be using to solve this?

$$y^{\prime\prime}+y=\tan x$$

-

The quick rigorous method I can think of is by variation of parameters.

The solution is given by $y = y_h + y_p$ where $y_h$ is the homogeneous part of the solution and $y_p$ is the particular solution.

The solution to the homogeneous part is $y_h(x) = c_1 \cos(x) + c_2 \sin(x)$.

$y_p$ is obtained by variation of parameters as follows:

The reason for writing it as a linear combination of $\cos(x)$ and $\sin(x)$ is that these two are the linearly independent solution to the homogeneous part.

$$y_p = a(x) \cos(x) + b(x) \sin(x)$$ $y_p' = a'(x) \cos(x) + b'(x) \sin(x) - a(x) \sin(x) + b(x) \cos(x)$.

Set $a'(x) \cos(x) + b'(x) \sin(x) = 0$ and hence $y_p' = - a(x) \sin(x) + b(x) \cos(x)$.

$y_p'' = - a'(x) \sin(x) + b'(x) \cos(x) - a(x) \cos(x) -b(x) \sin(x) = b'(x) \cos(x) - a'(x) \sin(x) - y_p$

Hence, we have $a'(x) \cos(x) + b'(x) \sin(x) = 0$ and $- a'(x) \sin(x) + b'(x) \cos(x) = \tan(x)$.

Solve for $a(x)$ and $b(x)$ from the two equations to get $b'(x) = \sin(x)$ and $a'(x) = -\frac{\sin^2(x)}{\cos(x)}$.

From which we get $b(x) = -\cos(x)$ and $a(x) = \sin(x) + \log \left( \left| \frac{\cos(x/2) - \sin(x/2)}{\cos(x/2) + \sin(x/2)} \right| \right)$

and plug it back in and simplify to get the particular solution as

$$y_p = \cos(x) \log \left( \left| \frac{\cos(x/2) - \sin(x/2)}{\cos(x/2) + \sin(x/2)} \right| \right)$$

The final solution is

$$y = c_1 \cos(x) + c_2 \sin(x) + \cos(x) \log \left( \left| \frac{\cos(x/2) - \sin(x/2)}{\cos(x/2) + \sin(x/2)} \right| \right)$$

EDIT: Taking a cue from Aryabhata's post, the Green's function for this equation (which is nothing but a 1D Helmholtz equation with unit wavenumber) is $G(x) = -i \frac{e^{i|x|}}{2}$ and hence the particular solution is $\int G(x-y) \tan(y) dy$

-
+1 A good answer, assuming that understanding was the point of the post. – Ross Millikan Mar 2 '11 at 6:25
Yep, I was able to find a'(x)=-sin(x)tan(x) and b'(x)=sin(x) and integrating for those gave me the right answer. – user6895 Mar 2 '11 at 6:32

The method below will solve equations of the form:

$$y'' + y = \frac{f'(x)}{\cos x}$$

First notice that $\displaystyle (h \cos x)'' = h'' \cos x - 2 h' \sin x - h \cos x$

Thus if $\displaystyle y = h \cos x$, then $\displaystyle y'' + y = h'' \cos x - 2h' \sin x$

Thus $\displaystyle (y'' + y')\cos x = h'' \cos^2 x - 2h' \sin x \cos x = (h' \cos^2 x)'$

Thus we get $$h' \cos^2 x = f(x) + A$$

And so

$$y = \cos x \int (f(x) + A)\sec^2 x \ \text{d}x$$

In your case, $\displaystyle f(x) = - \cos x$ and so

$$y = \cos x \ \int (A - \cos x) \sec^2 x \ \text{d}x = A\sin x - \cos x \ \log (\sec x + \tan x) + B \cos x$$

-
+1. Though I doubt if it is possible to come up with this idea in an timed test :). And this method should solve any right hand side i.e. $y'' + y = g$ then $y = \cos(x) \int ((\int g(y) \cos(y)dy) +A ) \sec^2(x) dx$ which is nothing but a Green's function sort of approach to solve the equation. – user17762 Mar 2 '11 at 7:35
@Sivaram: Yes, but once you know it... That is the case with almost every technique I would guess :-) – Aryabhata Mar 2 '11 at 7:36

It talks about "Variation of Parameters" which is what you need to use to solve it. A solution ends up being: $$y=-\cos(x)\ln\left(\sec(x)+\tan(x)\right).$$

All of this is covered in detail in the link. Also see http://en.wikipedia.org/wiki/Variation_of_parameters

-
Ah, thanks so much! – user6895 Mar 2 '11 at 6:29
Isn't what you gave a solution of $y'' + y = 1 + \tan x$? – Aryabhata Mar 2 '11 at 7:31
@Moron: You are right. Doesn't adding 1 just fix that? – Eric Naslund Mar 2 '11 at 7:38
Yes, but OP might be misled... – Aryabhata Mar 2 '11 at 7:44

If you feed d^2y/dx^2+y=tan(x) to Wolfram Alpha you get a solution.

-
Yeah, but that doesnt really help me learn how to do it... – user6895 Mar 2 '11 at 6:15
@woosh: Technically you did ask for the quickest way to solve the problem..., but I agree that doesn't help with learning! – Eric Naslund Mar 2 '11 at 6:35
@Eric: @woosh: I have a tendency to be very literal (or to take the cheap solution) But there are many ways to solve a problem. I hit the other answers with an upvote. – Ross Millikan Mar 2 '11 at 6:43