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Question
answer: \frac{8}{e^{8}}\\#\frac{da}{dt}=t + 4;a(2)=11.\
particular solution: \
to advance in the circuit, solve a(t)=-5.
Step1: Integrate the differential equation
We know that if \(\frac{dA}{dt}=t + 4\), then by the power rule of integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq- 1\)), we have \(A(t)=\int(t + 4)dt\).
Step2: Use the initial condition \(A(2) = 11\) to find \(C\)
Substitute \(t = 2\) and \(A(2)=11\) into \(A(t)=\frac{t^{2}}{2}+4t + C\).
So \(A(t)=\frac{t^{2}}{2}+4t + 1\)
Step3: Solve \(A(t)=-5\)
Set \(\frac{t^{2}}{2}+4t + 1=-5\).
Multiply through by \(2\) to get \(t^{2}+8t+2=-10\), or \(t^{2}+8t + 12=0\).
Factor the quadratic equation: \(t^{2}+8t + 12=(t + 2)(t+6)=0\)
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\(t=-2\) or \(t=-6\)