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

Question

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?

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

Answer 1

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 $\int G(x-y) \tan(y) dy$

Answer 2

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

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

Answer 3

Read this article: http://www.sosmath.com/diffeq/second/variation/variation.html
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

Answer 4

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