# 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)#