# second degree differential equation

Find all functions $f(x)$ such that

$$f''(x)+f(x)=\frac{1}{1+x^2}.$$

I would like to know if it's solvable and the solution/hints.

What I got :

$$2f(x)f(x)'+2f(x)'f(x)''=2\frac{1}{1+x^2}f'(x)$$ $$(f(x)^2)'+((f(x)')^2)'=2\frac{1}{1+x^2}f'(x)$$ $$f(x)^2+(f(x)')^2=\int 2\frac{1}{1+x^2}f'(x)dx .$$

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See my answer for a technique. Note that the particular soltion is in terms of special functions. –  Mhenni Benghorbal Jul 6 '14 at 5:05

## 2 Answers

If we use the method of Laplace transforms, denoting the Laplace transform of $f$ by $F$ and the Laplace transform of $1/(1+x^2)$ by $G$:

$$s^2 F(s) - s f(0) - f'(0) + F(s) = G(s)$$

and so

$$F(s) = \frac{G(s) + s f(0) + f'(0)}{s^2+1} \equiv \frac{H(s)}{s^2+1}$$

It is known (cf. Laplace transform tables) that the inverse Laplace transform of $\frac{1}{s^2+1}$ is $\sin(x)$. Then using the convolution property of the Laplace transform, we have $f(x)=h * \sin(x)$, where $h$ is the inverse Laplace transform of $H$ and $*$ denotes convolution. Linearity of the inverse Laplace transform gives $h(x) = \frac{1}{1+x^2} + \delta'(x) f(0) + \delta(x) f'(0)$, where $\delta$ is the Dirac delta. So we have the general solution

\begin{eqnarray*} f(x) & = & \int_0^x \sin(x-y) \left ( \frac{1}{1+y^2} + \delta'(y) f(0) + \delta(y) f'(0) \right ) dy \\ & = & f(0) \cos(x) + f'(0) \sin(x) + \int_0^x \frac{\sin(x-y)}{1+y^2} dy \end{eqnarray*}

This last integral cannot be expressed in terms of elementary functions.

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The whole intent was to calculate $y_{p}(\frac{\pi}{2})$. –  shooting-squirrel Jul 6 '14 at 5:34
That still can't be expressed in terms of elementary functions, though you can readily compute $\int_0^{\pi/2} \cos(y)/(1+y^2) dy$ with numerical methods. –  Ian Jul 6 '14 at 5:40

Yes this is solvable. Variation of parameters should do the trick.

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I have updated my question, some help with the problem would also be nice. –  shooting-squirrel Jul 6 '14 at 5:04
@dleggas. If I may ask, where are you arriving ? The result seems (at least to me) to be a monster ! Cheers. –  Claude Leibovici Jul 6 '14 at 5:14
I promise to show you after I get the computations done. –  shooting-squirrel Jul 6 '14 at 5:18