Differential Equation by Separation of Variables

[TranscendArcu]

Homework Statement

Solve the differential equation: y' + cos(x)y = cos(x)

The Attempt at a Solution

y' = cos(x)(1 -y)
\frac{dy}{1-y} = cos(x) dx
-ln|1-y| = sin(x) + C
\frac{1}{1-y} = e^{sin(x)} + C
1-y = \frac{1}{e^{sin(x)} + C}
y = \frac{-1}{e^{sin(x)} + C} + 1

Have I done this correctly?

 

[Dick]

No. e^(sin(x)+C) isn't the same as e^sin(x)+C.

 

[TranscendArcu]

So I know e^{sin(x) + C} = e^{sin(x)}e^C. But since e^C is just a constant, can't I just compress it into the constant C?

 

[Dick]

So I know e^{sin(x) + C} = e^{sin(x)}e^C. But since e^C is just a constant, can't I just compress it into the constant C?

Yes, but that makes it C*e^(sin(x)), doesn't it?

 

[TranscendArcu]

so I should write,

\frac{1}{1-y} = e^{sin(x)}C
1-y = \frac{1}{e^{sin(x)}C}, and C is totally arbitrary so write,
y = \frac{C}{e^{sin(x)}} + 1

 

[Dick]

so I should write,

\frac{1}{1-y} = e^{sin(x)}C
1-y = \frac{1}{e^{sin(x)}C}, and C is totally arbitrary so write,
y = \frac{C}{e^{sin(x)}} + 1

I think that works.