SOLUTION: The true solution at x=0.5 for y'= -xy at y(0)=1

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

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