Question 815508: The true solution at x=0.5 for y'= -xy at y(0)=1
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! 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
---------------------------------------
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 y = 0.88249690258456
|
|
|
