take Fourier transform of equation

Question

I was trying to solve this problem:

By taking the Fourier transform of the equation $$\frac{d^2\phi}{dx^2}-K^2\phi=f(x)$$ show that its solution, $\phi(x)$, can be written as $$\phi(x)=\frac{-1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty} \frac{e^{ikx}\tilde{f(k)}}{k^2+K^2}dk$$ where $\tilde{f(k)}$ is the Fourier transform of $f(x)$.

I tried this but I'm stuck..

Could someone show me how to take this Fourier transform?

$$\frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty} \frac{d^2\phi}{dx^2}-K^2\phi e^{-iwx} dx = \frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}f(x)e^{-iwx}dx $$

$$\frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty} \frac{d^2\phi}{dx^2} e^{-iwx}dx- \frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty} K^2\phi e^{-iwx} dx = \frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}f(x)e^{-iwx}dx $$

Answer 1

Hint:

1. Take the Fourier transform of both sides (you do not need to write down the integrals involved).
2. The left-hand side can be simplified by using: $$ \widehat{af+bg}=a\hat{f}+b\hat{g}$$ together with the rule for $\hat{g’}$.
3. Now you should have an expression like $$\hat{\phi}(t)(at^2+b)= \hat{f}(t)$$
4. Solve for $\hat{\phi}$ and take the inverse Fourier transform.