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

Question

Samantha Adkison · Accepted Answer

# y = ln(1/(A-e^sinx)) # is the General Solution 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)) #

Ezekiel Alexander · Answer

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

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)#

Axel Alvarez · Answer

undefined To solve the separable differential equation dy/dx = (cosx)e^(y+sinx), we can separate the variables and integrate both sides with respect to their respective variables. First, we divide both sides by (cosx)e^(y+sinx) to isolate dy/dx. This gives us:

dy / (cosx)e^(y+sinx) = dx

Next, we integrate both sides. The integral of dy / (cosx)e^(y+sinx) with respect to y involves a substitution. Let u = y + sinx, then du = dy. The integral becomes:

∫ (1 / (cosx)e^u) du

This integral can be simplified using trigonometric identities. Rewrite (1 / cosx) as secx, and the integral becomes:

∫ (secx / e^u) du

Now, we can integrate ∫ (secx / e^u) du. This integral involves integration by parts. Let dv = secx du and u = e^u. Then, differentiate u to find du, and integrate dv to find v. We have:

dv = secx du
du = e^u du

Integrating dv gives v = ln|secx + tanx| + C, where C is the constant of integration.

Now, we can apply integration by parts to find the integral of ∫ (secx / e^u) du. This gives us:

∫ (secx / e^u) du = ln|secx + tanx| + C

Now, we substitute back u = y + sinx:

ln|secx + tanx| = y + sinx + C

This is the solution to the differential equation dy/dx = (cosx)e^(y+sinx).