The general solution to the differential equation:
dy/dx +P(x)y = 0
is given by:
(1) y(x) = Ce^(-int_(x_0)^x P(t)dt)
Where x_0 is an arbitrary point x_o in I.
In fact we can see from direct substitution that:
dy/dx = -Ce^(-int_(x_0)^x P(t)dt) d/dx (-int_(x_0)^x P(t)dt)
dy/dx = -Ce^(-int_(x_0)^x P(t)dt) P(x) =-yP(x)
so that:
dy/dx + P(x)y = -P(x)y+P(x)y = 0
So we have that:
(a) For C = 0 we have the solution y(x) = 0.
(b) If we have y(x_0) = 0, we can choose x_0 as the lower bound of integration in (1), so that:
y(x_0) = Ce^(-int_(x_0)^(x_0) P(t)dt) = Ce^0 = C
Then:
y(x_0) = 0 => C=0
and the solution for C=0 is y(x) = 0
(c) If we have y_1(x_0) = y_2(x_0), we can choose x_0 as the lower bound of integration in (1), so that:
y_1(x_0) = C_1e^(-int_(x_0)^(x_0) P(t)dt) = C_1e^0 = C_1
y_2(x_0) = C_2e^(-int_(x_0)^(x_0) P(t)dt) = C_2e^0 = C_2
Then:
y_1(x_0) = y_2(x_0) => C_1=C_2
which means y_1(x) = y_2(x)
