How do you solve the separable differential equation #dy/dx=(cosx)e^(y+sinx)#?
2 Answers
# y = ln(1/(A-e^sinx)) # is the General Solution
Explanation:
We have:
# dy/dx = (cosx)e^(y+sinx) #
# dy/dx = (cosx)e^ye^sinx #
Which is a First Order Separable Differential Equation, which we can rewrite as:
# 1/e^ydy/dx = (cosx)e^sinx #
We can then "separate the variables" to get:
# int \ e^-y \ dy = int \ (cosx)e^sinx \ dx#
Which we can directly (and easily) integrate to get:
# - e^-y = e^sinx + B #
# :. e^-y = A-e^sinx #
# :. -y = ln(A-e^sinx) #
# :. y = -ln(A-e^sinx) #
# :. y = ln(1/(A-e^sinx)) #
Explanation:
To "separate" a differential equation means to move all the terms with
We treat
To separate this, we also need to split up
#dy/dx=(cosx)e^(y+sinx)#
#dy=(cosx)e^y(e^sinx)dx#
#dy/e^y=(cosx)e^sinxdx#
Now we can integrate both sides:
#inte^-ydy=inte^sinx(cosx)dx#
For the left-hand side, use the substitution
#-inte^-y(-dy)=inte^sinxcosxdx#
#-inte^u=inte^sinxcosxdx#
#-e^u=inte^sinxcosxdx#
#-e^-y=inte^sinxcosxdx#
Following a similar process on the right, let
#-e^-y=inte^tdt#
#-e^-y=e^sinx+C#
Solving for
#e^-y=-e^sinx+C#
#-y=lnabs(C-e^sinx)#
#y=-lnabs(C-e^sinx)#