# Separable Differential Equation

## Homework Statement

Solve the differential equation:
(ex+1)cosy dy + ex(siny +1)dx=0 y(0)=3

none

## The Attempt at a Solution

(ex+1)cosy dy + ex(siny +1)dx=0
(ex+1)cosy dy =- ex(siny +1)dx
cosy/(siny+1)dy=-ex/(ex+1)dx
∫cosy/(siny+1)dy=-∫ex/(ex+1)dx
using u sub on both the left and right for u=siny+1 on left and u=(ex+1) on right yeilds
∫1/u du=-∫ 1/u du
ln|siny +1|= -ln|ex+1| +c
exponentiating both sides ( this is where I think I messed up)
siny +1= -x-1 +c

Mark44
Mentor

## Homework Statement

Solve the differential equation:
(ex+1)cosy dy + ex(siny +1)dx=0 y(0)=3

none

## The Attempt at a Solution

(ex+1)cosy dy + ex(siny +1)dx=0
(ex+1)cosy dy =- ex(siny +1)dx
cosy/(siny+1)dy=-ex/(ex+1)dx
∫cosy/(siny+1)dy=-∫ex/(ex+1)dx
using u sub on both the left and right for u=siny+1 on left and u=(ex+1) on right yeilds
∫1/u du=-∫ 1/u du
ln|siny +1|= -ln|ex+1| +c
exponentiating both sides ( this is where I think I messed up)
siny +1= -x-1 +c
Yes, it's messed up.
Use the ideas that a * ln(b) = ln(ba) and that exp(m + n) = exp(m) * exp(n)

So all of my work is fine up until the last step?

Mark44
Mentor
I don't see anything wrong, but you shouldn't use u for both of your substitutions -- maybe v or w for the other one.

yea, I figured that was bad notation. Anyways, for the next step, is the left side of the equation is right? sin(y) +1

as far as the right side and applying the rules you gave me. The right side would look something like this before the exponentiating:

-(x+1)ln|e| +c

Mark44
Mentor
You have a mistake that I didn't notice before.
ln|siny +1|= -ln|ex+1| +c
On the right side it should be ##-\ln(e^x + 1) + C##. You had ##-\ln|e^{x + 1}| + C##.
Also note that I didn't use absolute values, since ##e^x > 0## for all real x, so ##e^x + 1 > 1 > 0## for all real x.

You don't need absolute values on the left side, either, since ##\sin(y) + 1 \ge 0## for all real y.