Solve First order linear differential equation, initial conditions

[NCyellow]

Homework Statement

The problem is given as follows:
Solve
dy/dt + y = 0.5, y(t=0)=1

Homework Equations

The Attempt at a Solution

I separate the y terms from the t terms, which gives me
dy(-y+0.5)=dt
I integrate both sides to get
-ln(-y+0.5)=t+C
C is the constant, I combine the constants from both sides to one value.
Multiplying both sides by the negative,
ln(-y+0.5)=-t-C
Now i e both sides
-y+0.5=e^(-t-C)
Therefore I can simplify to y=e^(-t-C)+0.5, which is my solution
Since y(0)=1,
e^(-C) = 0.5
I don't know what exactly I am supposed to do with that...
Was my answer correct?
Please advise, Thank you!

 

[latentcorpse]

surely you mean $e^{-C}=1.5$

 

[NCyellow]

surely you mean $e^{-C}=1.5$

Wrote the signs wrong, fixed now.

 

[HallsofIvy]

Homework Statement

The problem is given as follows:
Solve
dy/dt + y = 0.5, y(t=0)=1

Homework Equations

The Attempt at a Solution

I separate the y terms from the t terms, which gives me
dy(-y+0.5)=dt

dy/(-y+0.5)= dt

I integrate both sides to get
-ln(-y+0.5)=t+C
C is the constant, I combine the constants from both sides to one value.

Multiplying both sides by the negative,
ln(-y+0.5)=-t-C
Now i e both sides
-y+0.5=e^(-t-C)
Therefore I can simplify to y=e^(-t-C)+0.5, which is my solution
Since y(0)=1,
e^(-C) = 0.5
I don't know what exactly I am supposed to do with that...
Was my answer correct?
Please advise, Thank you!

$-y+0.5= e^{-t-C}$ is NOT equivalent to $y= e^{-t-C}+0.5$.

Subtracting 0.5 from each side of $-y+0.5= e^{-t-C}$ gives $-y= e^{-t-C}- 0.5$ and then, multiplying both sides by -1, $y= 0.5- e^{-t-C}$.

Now, taking y= 1 when t= 0, $1= 0.5- e^{-C}$ so $e^{-C}= 0.5$ still.
And $e^{-t-C}= e^{-t}e^{-C}$. Since $e^{-C}= 0.5$, $e^{-t-C}= 0.5e^{-t}$ and your solution is $y= 0.5- 0.5e^{-t}$