Solve the differential equation # dy/dx + y = 0# ?

2 Answers
May 14, 2017

# y = Ae^(-x) #

Explanation:

We can rewrite the equation:

# dy/dx + y = 0#

as:

# dy/dx = -y => 1/ydy/dx=-1`#

Which is a First Order linear separable Differential Equation, so we can "separate the variables" to get:

# int \ 1/y \ = int \ -1 \ dx #

Which we can integrate to get:

# ln |y| = -x + c #

Taking Natural logarithms we then get:

# |y| = e^(-x + c) #

As #e^x > 0 AA x in RR# we then get:

# y = e^(-x + c) #
# \ \ = Ae^(-x) #

We can easily verify the solution:

# y = Ae^(-x) => y' = -Ae^(-x)#
# y' + y = -Ae^(-x) + Ae^(-x) = 0 \ \ \ QED#

May 14, 2017

#y = Ce^-x; C > 0#

Explanation:

Given: #dy/dx + y = 0#

Subtract y from both sides of the equation:

#dy/dx = -y#

We shall use the separation of variables method.

To do this, we multiply both sides by #dx# and divide both sides by #y#:

#dy/y = -dx#

Please observe that we have "separated" all of the things to with y on the left and all of the things to do with x on the right.

We can integrate both sides:

#intdy/y = -intdx#

The left side becomes the natural logarithm and the right becomes x plus an arbitrary constant:

#ln|y| = -x+c#

Eliminate the logarithm by making both sides exponents of e:

#e^(ln|y|) = e^(-x+c)#

By the definition of the inverse function of a function the left side becomes y:

#y = e^(-x+c)#

We can write the some of two exponents as a product:

#y = (e^c)(e^-x)#

e to an arbitrary constant is just another constant but is can only be positive, because that is the range of e:

#y = Ce^-x; C > 0 #

Check the solution:

#dy/dx = -Ce^-x#

Substitute y and #dy/dx# into the original equation:

#-Ce^-x + Ce^-x= 0#

#0= 0#

This checks.