Initial value $\left ( \frac{dy}{dt} \right )+3y=11$, $y(0)=1$ I have never done an initial value problem, and would like some help on how to start this please.
The first step is to solve the ODE:

$$y'+3y=11\xrightarrow{\text{Using an appropriate integrating factor }~\mu(x)=\exp(3x)}d\left(e^{3x}y\right)=11e^{3x}$$ then follow the way posted in @Shaaeh'spost.
dy/dt=11-3y

=>dy/(11-3y)=dt     (provided y!=11/3)

=>ln(y-11/3)=-3t+C    (integrating both sides where c is constant of integartion)

=>y=(11/3)+exp(-3t)*C1  (where C1=exp(C) which is another constant)

it is given that y(0)=1 i.e., when t=0; y=1;

now we need to find the exact equation and therefore we substitute the value of t=0 and y=1 in the equation.

we get

C1=-8/3;

therefore the exact solution is y=(11/3)+(-8/3)exp(-3*t)