Find a continuous solution satisfying: $$ \frac{dy}{dx} +y= f(x) $$ Where $$ f(x) = \begin{cases} 1 &\text{ for } 0 < x < 1, \\ 0 & \text{ for } x > 1 \end{cases} $$ with the initial condition $y(0)=0$. With use of the Fourier Transform
I understand the definition of the Fourier transform, and have found it for $f(x)$:
$F(x) = \frac{i}{k}\cdot (1 -e^{-ik})$
But I'm not sure how to then solve the ODE using this?
Can someone please help me with the next steps to find a solution please.
On
Proceeding informally, suppose $y, y'$ have Fourier transforms, then, depending on your definition of the Fourier transform we have $(1+ i \omega) \hat{y}(\omega) = {1 \over i \omega}(1-e^{-i \omega})$ and so $\hat{y}(\omega) = {1 \over i \omega (1 + i \omega)} (1-e^{-i \omega}) $.
A little bit of work shows that $y(x) = u(x)(1-e^{-x}) - u(x-1)(1-e^{-(x-1)})$.
In this case, for $x \le 0$ we know that there is a unique solution to the differential equation ($x \mapsto 0)$ and for
Denote the fouriertransform of $y$ for $\hat{y}$ then the fourierstransform of $y'$ is $2\pi i t \hat{y}$, so your ODE becomes
$$ 2\pi i t \hat{y}(t) + \hat{y}(t) = - \frac{i- ie^{-2\pi i t}}{2t\pi} $$
Where $\int_{0}^{1} e^{-it2\pi x} dx = - \frac{i- ie^{-2\pi i t}}{2t\pi}$
Now you can solve for $\hat{y}$ and use the inverse fouriertransform.