It seems you are asking too much. The solution is uniquely determined by the boundary conditions. Once found a solution you can check if it is consistent with the value you have at ##(x_i,y_i)##.
Hello I would like to check my reasoning about solutions of first order PDE. I've spell out (almost) all details.
I'll consider the following problem: find ##u=u(t,x)## s.t. :
$$ \partial_t u(t,x) + a(x) \cdot \nabla u(x) =0 \qquad \qquad u(0,x) = u_0(x)$$
say with smooth coefficient and...
Gotcha!
$$\frac{d h(t)}{d t} =\frac{d}{ d t}\int_0^t f(t-\tau) g(\tau) d \tau $$
and using Leibniz integral rule
$$ \frac{d h(t)}{d t} = f(0)g(t) + \int_0^t \frac{d f(t-\tau)}{ d t} g(\tau) d \tau $$
The laplace transform of which is
$$ s H(s) = G(s) f(0) + G(s)(s F(s) - f(0) ) = s F(s)G(s)...
Prelude
Consider the convolution h(t) of two function f(t) and g(t):
$$h(t) = f(t) \ast g(t)=\int_0^t f(t-\tau) g(\tau) d \tau$$
then we know that by the properties of convolution
$$\frac{d h(t)}{d t} = \frac{d f(t)}{d t} \ast g(t) = f(t) \ast \frac{d g(t)}{d t}$$
Intermezzo
We also know that...
Thanks, following your hint this is I would go:
if ##y_{\pm}## solve ##y'_{\pm}(x)+a(x)*y_{\pm}(x)=0## in ## \pm x> 0## then ##y_{\pm} = C_{\pm} \, e^{-A_{\pm}(x)} ## and a solution is given by
y(x) = y_{-}(x) + H(x) (u_{+}(x) - u_{-}(x))
where
A_{+}(x) = \int_{0^{+}}^{x} a(z) dz...
I'm trying to solve the following equation (even if I'm not sure if it's well posed)
\partial_{x} \, y(x) + a(x)\, y(x) = \delta(x)
with ##\quad \lim_{x \rightarrow \pm \infty}y(x) = 0##
It would be a classical first order ODE If it were not for the boundary conditions and the Dirac...