How do you solve the differential equation #y'=e^(-y)(2x-4)#, where #y5)=0# ?

1 Answer
Sep 8, 2014

This is an example of a separable equation with an initial value.
#{dy}/{dx}=e^{-y}(2x-4)#
#y(5)=0#

by multiplying by #e^y# and by #dx#,
#Rightarrow e^ydy=(2x-4)dx#
by integrating,
#Rightarrow inte^ydy=int(2x-4)dx#
#Rightarrow e^y=x^2-4x+C#
by taking the natural log,
#Rightarrow y=ln(x^2-4x+C)#

Now, we need to find #C# using #y(5)=0#.
#y(5)=ln((5)^2-4(5)+C)=ln(5+C)=0#
#Rightarrow 5+C=1 Rightarrow C=-4#

Hence, the solution is #y=ln(x^2-4x-4)#.