Question

solve the initial value problem.

\\( \frac { d y } { d t } + 2 y = 11, y ( 0 ) = 1 \\)

the solution is \\( y = \\)

Explanation:

Step1: Find the integrating factor

For a first - order linear differential equation \(\frac{dy}{dt}+P(t)y = Q(t)\), here \(P(t)=2\), \(Q(t) = 11\).
The integrating factor \(\mu(t)=e^{\int P(t)dt}\).
Since \(\int 2dt=2t\), then \(\mu(t)=e^{2t}\).

Step2: Multiply the differential equation by the integrating factor

Multiply \(\frac{dy}{dt}+2y = 11\) by \(e^{2t}\), we get \(e^{2t}\frac{dy}{dt}+2e^{2t}y = 11e^{2t}\).
By the product rule \((uv)^\prime=u^\prime v+uv^\prime\), where \(u = y\) and \(v=e^{2t}\), the left - hand side is \(\frac{d}{dt}(ye^{2t})\).
So \(\frac{d}{dt}(ye^{2t})=11e^{2t}\).

Step3: Integrate both sides

Integrate \(\frac{d}{dt}(ye^{2t})=11e^{2t}\) with respect to \(t\).
\(\int\frac{d}{dt}(ye^{2t})dt=\int 11e^{2t}dt\).
Using \(\int e^{at}dt=\frac{1}{a}e^{at}+C\) (\(a = 2\)), we have \(ye^{2t}=\frac{11}{2}e^{2t}+C\).
Then \(y=\frac{11}{2}+Ce^{-2t}\).

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

Substitute \(t = 0\) and \(y=1\) into \(y=\frac{11}{2}+Ce^{-2t}\).
\(1=\frac{11}{2}+C\).
Solve for \(C\): \(C=1-\frac{11}{2}=-\frac{9}{2}\).

Answer:

\(y=\frac{11}{2}-\frac{9}{2}e^{-2t}\)