How do you find the particular solution to #y(x+1)+y'=0# that satisfies y(-2)=1?
1 Answer
Nov 11, 2016
# y = e^(-1/2x^2-x)#
Explanation:
We have
# y(x+1)+y'=0 #
# y' = -y(x+1) #
# :. int 1/y dy = int-(x+1) dx#
Integrating gives us;
# ln y = -1/2x^2-x+C#
And# y(-2)=1 => ln1=(-1/2)(4)-(-2)+C#
# :. 0=-2+2+C=>C=0#
So the particular solution is
# ln y = -1/2x^2-x#
# :. y = e^(-1/2x^2-x)#
