Question

Let y(x) satisfy the ordinary differential equation

d^2y/dx^2 - (a^2)y = f(x)

where a>0, and y(x) tends rapidly to zero as x tends to +/- ∞. Show that the Fourier transform ŷ(k) of y(x) is given by

ŷ(k) = - f^(k)/(k^2+a^2)

(in the line above f^(k) is supposed to be f-hat(k), sorry couldn't find the right symbol)

Answer 1

Question 29585

Let us take the Fourier transform from both sides of equation.

To do this, one should use the property of Fourier transform $(\hat{y}^n) = (ik)^n\hat{y} (k)$ , where $y^{n}$ denotes n-th derivative.

For equation $y^{\prime \prime}(x) - a^{2}y(x) = f(x)$ , Fourier transform will be $(ik)^{2}\hat{y} (k) - a^{2}\hat{y} (k) = \hat{f} (k)$ .

Solving this equation for $\hat{y} (k)$ gives $\hat{y} (k) = \frac{-\hat{f}(k)}{k^2 + a^2}$ .