I am stumped on solving the following diff eq:

$\displaystyle \frac{\,dy}{\,dx}=\frac{y}{3+t}$ with the initial condition $\displaystyle y(0)=1$

I have figured the general solution to be $\displaystyle y=3+t+C$,

from $\displaystyle \ln\left|y\right|=\ln\left|3+t\right|+C$,

but it doesn't yield the correct result: $\displaystyle y=\frac{1}{3}(3+t)$

I keep getting $\displaystyle y=1+t$.

Please, what am I doing wrong?

Thank you,

Jen