How do you solve the separable differential equation #dy/dx=(cosx)e^(y+sinx)#?

2 Answers
Mar 13, 2017

# 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)) #

Mar 13, 2017

#y=-lnabs(C-e^sinx)#

Explanation:

To "separate" a differential equation means to move all the terms with #y# to one side of the equation, and all the terms with #x# to the other side of the equation.

We treat #dy/dx# as a division problem, so we can move the #dx# to the other side of the equation, leaving just #dy#.

To separate this, we also need to split up #e^(y+sinx)# as #e^y(e^sinx)#.

#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 #u=-y#, implying that #du=-dy#.

#-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 #t=sinx# so #dx=cosxdx#.

#-e^-y=inte^tdt#

#-e^-y=e^sinx+C#

Solving for #y#:

#e^-y=-e^sinx+C#

#-y=lnabs(C-e^sinx)#

#y=-lnabs(C-e^sinx)#