1. ## I'm having some algebraic difficulties: Help Please! Also Integration

1) Could you please demonstrate the algebra going from

1/4 ln (u) = sinx + c to [u]=Ae^(4sin(x)) and How is the second function similar to ln(y^(4)+1)= C+ 4sin(x)

I would like to know how 4 ended up being part of the exponent. (I forgot how to solve)

2) How would you simplify e^(y) = 3e^(2x)+C by solving for y?
y= ?
(The back of my book has the final answer as y= ln(3e^(2x)-2)

I do not understand where the -2 comes from. Here's what the original problem said in case you need to refer to it:

*Find an explicit particular solution of the initial problem :
dy/dx= 6e^(2x-y) , y(0)=0.
(I know the Physics Maestro helped me with this question but the teacher has not tought us that way because he is trying to make it easier for us because it's summer!

2. *Find an explicit particular solution of the initial problem :
dy/dx= 6e^(2x-y) , y(0)=0.
(I know the Physics Maestro helped me with this question but the teacher has not tought us that way because he is trying to make it easier for us because it's summer!
Rewrite as $y'=6e^{2x}e^{-y}$

Separate variables: $\frac{y'}{e^{-y}}=6e^{2x}$

Integrate: $\int{e^{y}}dy=6\int{e^{2x}}dx$

$e^{y}=3e^{2x}+C$

$y=ln(3e^{2x}+C)$

IC is y(0)=0: $0=ln(3e^{2(0)}+C)$

C=-2

$y=ln(3e^{2x}-2)$