I have this PDE: $$ \, u_t - u_{xx}+u_x+u=0, $$ with IC $ \, u(x,0) = f(x) $, where $ x \in \mathbb{R},\, t > 0 $.
Using Fourier transform i was able to find solution: $$ u(x,t)=\frac{1}{e^{t}\sqrt{4\pi t}} \int_{-\infty}^{\infty } f(s) e^\frac{-(x-(s+t))^2}{4t} ds $$
Now I am trying to verify that this solution is correct by substituting it into the original equation. Assuming that the derivative and the integral are interchanged, I count the individual terms. $u_t$ and $u_{xx}$ cancel out, but the remaining two terms don't. So I'm investigating whether $ u_x = - u $, but I don't know how to do it. I calculated that $u_x$ is $$ u_x(x,t) = \frac{1}{e^{t}4t\sqrt{\pi t}} \int_{-\infty}^{\infty } f(s) e^\frac{-(x-(s+t))^2}{4t} \cdot(s+t-x) ds $$
EDIT:
Ok, so I made a mistake. I grouped the integrals wrong. My solution is correct and satisfies equation.
$$ u_t - u_{xx}+u_x+u=0 $$
Write $u$ as $$ u(x,t) = \int_{-\infty}^{\infty} \hat{u}(\omega,t) e^{i 2 \pi \omega x} \: d\omega $$
Then $$u_{t} = \int_{-\infty}^{\infty} \hat{u}_{t}(\omega,t) e^{i 2 \pi \omega x} \: d\omega $$
and
$$ u_{x} = (i 2 \pi \omega) \int_{-\infty}^{\infty} \hat{u}(\omega,t) e^{i 2 \pi \omega x} \: d\omega $$
$$ u_{xx} = -( 2 \pi \omega)^{2} \int_{-\infty}^{\infty} \hat{u}(\omega,t) e^{i 2 \pi \omega x} \: d\omega $$
pluggin to the equation you get
$$ \int_{-\infty}^{\infty} \left[ \hat{u}_{t}(\omega,t) + ((2 \pi \omega)^{2} + i2\pi \omega +1) \hat{u}(\omega,t) \right] e^{i 2 \pi \omega x} \: d\omega = 0$$
So we have the "transformed" equation:
$$ \left[ \hat{u}_{t}(\omega,t) + ((2 \pi \omega)^{2} + i2\pi \omega +1) \hat{u}(\omega,t) \right] = 0$$
This has the form
$$ \hat{u}_{t} = -K(\omega)\hat{u} $$
which has solution $\hat{u} = C(\omega) e^{-K(\omega)t}$.
The transformed initial condition is $\hat{u}(\omega,0) = \hat{f}$. So $C(\omega)=\hat{f}$.
Hence
$$\hat{u} = \hat{f} e^{-((2 \pi \omega)^{2} + i2\pi \omega +1)t}$$
Then $u(x,t)$ must be the inverse of $\hat{u}$. Is this how you got your solution $u(x,t)$???
Btw, here is a video that may be helpful. About Fourier transform on PDE: https://youtu.be/_eq8rUr1QIg