Question 815508
y'= -xy


dy/dx= -xy


dy= -xy*dx


dy/y= -x*dx


int( dy/y )= int( -x*dx ) ... integrating both sides


ln( |y| ) = (-1/2)*x^2 + C ... don't forget the " + C " part


|y| = e^[(-1/2)*x^2 + C]


|y| = e^((-1/2)*x^2)*(e^C)


|y| = e^C * e^((-1/2)*x^2)


y = +-e^C * e^((-1/2)*x^2)


Because e^C and e^((-1/2)*x^2) are NEVER negative, this means that the expression e^C * e^((-1/2)*x^2) is NEVER negative.


So that fact, along with the fact that y(0) = 1, which is positive, means that we're going to focus on the "plus" and ignore the "minus"


Basically, we now have y = e^C * e^((-1/2)*x^2)


Plug in x = 0 and y = 1, from the initial condition y(0) = 1, then solve for C


y = e^C * e^((-1/2)*x^2)


1 = e^C * e^((-1/2)*0^2)


1 = e^C * e^((-1/2)*0)


1 = e^C * e^(0)


1 = e^C * 1


1 = e^C


e^C = 1


C = ln(1)


C = 0


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So


y = e^C * e^((-1/2)*x^2)


turns into


y = e^0 * e^((-1/2)*x^2)


y = 1 * e^((-1/2)*x^2)


y = e^((-1/2)*x^2)


which is the solution that satisfies both the differential equation y' = -xy and the initial condition y(0) = 1


To find the value of y at x = 0.5, plug it in and evaluate.


y = e^((-1/2)*x^2)


y = e^((-1/2)*(0.5)^2)


y = e^((-1/2)*0.25)


y = e^(-0.125)


y = 0.88249690258456 (this is approximate)


So when x = 0.5, y is approximately <font color="red">y = 0.88249690258456</font>