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.
 A: 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.
A: The answer is 
$$y(t)=c_1e^{-3t}+\frac{11}{3}$$
but $c_1$ is unknown. so we substitute $t=0$ and $y=1$ :
$$1=c_1+\frac{11}{3}$$
$$c_1=-\frac{8}{3}$$  so $$y(t)=-\frac{8}{3}e^{-3t}+\frac{11}{3}$$
A: $$\frac{dy}{dt} + 3y = 11\ \ ... (1)
$$
homogeneous form is
$$
\frac{dy}{dt} + 3y = 0\\
\int \frac{dy}{y} = -3\int dt\\
\log{|y|} = -3t + C\\
y = \exp(-3t + C)\\
y = C\exp(-3t)\\
$$
variation of Parameters
C => u(t)
$$
y = u(t)\exp(-3t)\ \ ...(2)\\
\frac{dy}{dt} = \frac{du}{dt}\exp(-3t) - 3u\exp(-3t)\ \ ...(3)
$$
(2), (3) => (1)
$$ \frac{du}{dt}\exp(-3t)-3u\exp(-3t) + 3u\exp(-3t) = 11\\
\frac{du}{dt}\exp(-3t) = 11\\
\int du = 11\int \exp(3t)dt\\
u = \frac{11}{3}\exp(3t) + C
$$
therefore,
$$
y = (\frac{11}{3}\exp(3t) + C)\exp(-3t)\\
y = C\exp(-3t) + \frac{11}{3}
$$
now, y(0) = 1
$$
y(0) = 1 = C + \frac{11}{3}\\
C = -\frac{8}{3}
$$
so,
$$
y = -\frac{8}{3}\exp(-3t) + \frac{11}{3}
$$
A: Two simple methods come to mind:
1) Rearrange, divide through, and integrate
$\dfrac{dy}{dt} = 11 - 3y$
Which can be rearranged to:
$\int^{y(t)}_{y(0)}\dfrac{ds}{11 - 3 s} = \int^{t}_{0}du$
where the limits of integration are from your initial value.
This works as $y$ and $t$ are effectively 'separated', as there are no $t$ terms in the equation. If there were, and $y$ and $t$ weren't separable, then the below method would need to be used.

2) Integrating factor
You have (in general terms):
$\dfrac{dy}{dt} + f(t) y = g(t)$.
View this as:
$\dfrac{d}{dt}(h(t)y) = h(t)\dfrac{dy}{dt} + h'(t)y = h(t)g(t)$,
i.e. $h(t)$ multiplying the original equation, where
$h'(t) \equiv \dfrac{dh}{dt}$,
such that (after matching the functions multiplying $y$):
$f(t) = \dfrac{h'(t)}{h(t)} = \dfrac{d\ln{h}}{dt}$.
This leads to an expression for the integrating factor (one needn't worry about the constant of integration as this results in multiplying the whole equation by a scalar):
$h(t) = \exp\lgroup{\int^{t}f(s)ds}\rgroup$.
This leads to:
$\dfrac{d}{dt}(h(t)y) = h(t)g(t)$
$h(t)y(t) - h(0)y(0) = \int^{t}_{0}h(s)g(s)ds $
And in your case:
$f(t) = 3$
$g(t) = 11$
$h(t) = e^{3t}$.
A: Multiplying both sides through by the integrating factor $e^{\int 3\,\text{d}t}=e^{3t}$ and integrating on both sides:
$e^{3t}y=\int e^{3t}\cdot 11\,\text{d}t$
$y=\frac{\frac{11}{3}e^{3t}+C}{e^{3t}}=\frac{11}{3}+Ce^{-3t}$
Initial value: $y(0)=1$:
$1=\frac{11}{3}+Ce^{-3\cdot0} \Longleftrightarrow C=\frac{1-\frac{11}{3}}{e^{-3\cdot0}}=-\frac{8}{3}$
The solution becomes:
$\Large{y=\frac{11}{3}-\frac{8}{3}e^{-3t}}$
A: 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)
