Separable Differential Equation

[Dusty912]

Homework Statement

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

Homework Equations

none

The Attempt at a Solution

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

 

[Mark44]

Homework Statement

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

Homework Equations

none

The Attempt at a Solution

(e^x+1)cosy dy + e^x(siny +1)dx=0
(e^x+1)cosy dy =- e^x(siny +1)dx
cosy/(siny+1)dy=-e^x/(e^x+1)dx
∫cosy/(siny+1)dy=-∫e^x/(e^x+1)dx
using u sub on both the left and right for u=siny+1 on left and u=(e^x+1) on right yeilds
∫1/u du=-∫ 1/u du
ln|siny +1|= -ln|e^x+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(b^a) and that exp(m + n) = exp(m) * exp(n)

 

[Dusty912]

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

 

[Mark44]

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.

 

[Dusty912]

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]

You have a mistake that I didn't notice before.

ln|siny +1|= -ln|e^x+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.