<s>[INST] 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.
 [/INST] $$\frac{dy}{dt} + 3y = 11\ \ ... (1)
$$
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,
$$
y = -\frac{8}{3}\exp(-3t) + \frac{11}{3}
$$
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