Question

solve the initial value problem.
\\( \frac { d y } { d t } + 3 y = 11, y ( 0 ) = 1 \\)
the solution is \\( y = \\)

Explanation:

Step1: Identify the integrating factor

For the linear differential equation \(\frac{dy}{dt}+P(t)y = Q(t)\), here \(P(t)=3\), \(Q(t) = 11\). The integrating factor \(\mu(t)=e^{\int P(t)dt}\).
\(\int 3dt=3t\), so \(\mu(t)=e^{3t}\).

Step2: Multiply the equation by the integrating factor

Multiply \(\frac{dy}{dt}+3y = 11\) by \(e^{3t}\):
\(e^{3t}\frac{dy}{dt}+3e^{3t}y = 11e^{3t}\).
By the product rule \((uv)^\prime=u^\prime v+uv^\prime\), where \(u = y\), \(v=e^{3t}\), the left - hand side is \(\frac{d}{dt}(ye^{3t})\). So \(\frac{d}{dt}(ye^{3t})=11e^{3t}\).

Step3: Integrate both sides

Integrate \(\frac{d}{dt}(ye^{3t})=11e^{3t}\) with respect to \(t\):
\(ye^{3t}=\int 11e^{3t}dt\).
Using \(\int ae^{bt}dt=\frac{a}{b}e^{bt}+C\) (\(a = 11\), \(b = 3\)), we have \(\int 11e^{3t}dt=\frac{11}{3}e^{3t}+C\).
So \(ye^{3t}=\frac{11}{3}e^{3t}+C\), and \(y=\frac{11}{3}+Ce^{-3t}\).

Step4: Use the initial condition \(y(0) = 1\)

Substitute \(t = 0\) and \(y=1\) into \(y=\frac{11}{3}+Ce^{-3t}\):
\(1=\frac{11}{3}+C\).
Solve for \(C\): \(C=1-\frac{11}{3}=-\frac{8}{3}\).

Answer:

\(y=\frac{11}{3}-\frac{8}{3}e^{-3t}\)