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)#
